Limits of complex functions, examples of solutions. Limit of sequence and function. Limit theorems

2011 Viosagmir I.A. Function limit 2011 Higher mathematics for dummies. Function limit [email protected] Higher mathematics for dummies. Limit of a function 2011 1 Limit of a function Introduction Well... I welcome you to my first book dedicated to the limits of a function. This is the first part of my upcoming series “higher mathematics for dummies”. The title of the book should already tell you a lot about it, but you may completely misunderstand it. This book is dedicated not to “dummies,” but to all those who find it difficult to understand what professors do in their books. I'm sure you understand me. I myself was and am in such a situation that I am simply forced to read the same sentence several times. This is fine? I think no. So what makes my book different from all the others? Firstly, the language here is normal, not “abstruse”; secondly, there are a lot of examples discussed here, which, by the way, will probably be useful to you; thirdly, the text has a significant difference from each other - the main things are highlighted with certain markers, and finally, my goal is only one - your understanding. Only one thing is required from you: desire and skills. “Skills?” - you ask. Yes! Skills and. In general, it is recommended to keep a separate notebook of about 65 sheets, and write everything in it. Everything that is written in this book. The result will be impressive, I promise you that. It is also better to use multi-colored markers. Well, gentlemen... I want to wish you success and understanding. If you finish this book, you will be able to do a lot!!! There will be some notations throughout my book. I highly recommend following them. - be sure to learn! - It is recommended to try to do it yourself. - You don’t have to teach it, but you need to understand it! Higher mathematics for dummies. Limit of a function 2011 2 Contents Limit of a function at a point………………………………………………………………………………………….3 Theorems about limits…………………………………………………………………………………………………………..13 One-sided limits………… …………………………………………………………………………………..14 Limit at →∞…………………………… ……………………………………………………………………..17 Infinitely large functions………………………………………………………. …………………………………………………25 Graphs of elementary functions………………………………………………………………………………………… …………..26 Continuity of a function at a point………………………………………………………………………………….31 Continuity of a complex function……… …………………………………………………………………………………..33 Classification of break points……………………………………………………………… ………………………………………………………36 Continuity of elementary functions………………………………………………………………………………41 The first remarkable limit…………………………………………………………………………………..42 The second remarkable limit…………………………… ………………………………………………………..47 Briefly about Maple……………………………………………………… …………………………………………………..52 Comparison of infinitesimal functions……………………………………………………………… …………. .55 Properties of the symbol “o small”…………………………………………………………………………………………………………..60 Asymptotic formulas……… ……………………………………………………………………………………64 L'Hopital's Rule……………………………………………………… ………………………………………………………………………………72 Taylor series expansion. Part 1………………………………………………………………………………..80 Taylor series expansion. Part 2………………………………………………………………………………..88 Higher mathematics for dummies. Function limit 2011 3 Chapter 1. Function limit. Let it be a numerical variable, the area of ​​its change. If each number ∈ is associated with a certain number, then they say that a function is defined on the set and write. I hope this is clear to you, but I’ll explain just in case. The set in this case is a plane consisting of two coordinate axes – 0X and 0Y. You should have known this since school. If you forgot this, open class 7 - 8 and repeat. For example, in Fig. 1 shows the function. The 0X and 0Y axes form the area of ​​its change. We see perfectly in Fig. 1, how the function behaves. In this case, they say that a function is defined on the set. The set of all partial values ​​of a function is called a set of values. In other words, the set of values ​​is the interval along the OY axis where the function is defined. For example, consider Fig. 1. – from here it is immediately clear that 0, because 0. This is clearly visible in the figure. In this case, the range of values ​​is 0;∞. Remember, we look at many values ​​by 0Y! The set of all is called the domain of definition. We draw a conclusion from the previous considerations and understand that we look at the set of definitions by 0. In our case, ODZ = ∞;∞. A point ∈ or is called a limit point of a set if in any neighborhood of the point there are points of the set different from. I won’t add anything here. And so everything is clear. We can only add that in our case the limit point of the set is the domain of definition of the function. Contents: 1) Limit of a function at a point 2) Theorems about limits 3) One-sided limits 4) Limit, at →∞ 5) Infinitely large functions 6) Graphs of elementary functions 1. Limit of a function at a point. Rice. 1 independent variable (argument). domain of definition of the function. partial value of a function at a point. Higher mathematics for dummies. Limit of a function 2011 4 So, before defining it, let me explain in general terms what the limit of a function is. The number b that the function tends to as x tends to the number is called the limit of the function. This is how it’s all written: lim → For example, . We need to find out what the function tends to (is not equal!) as →2. First, let's write down the limit: lim → lim → Now it's time to look at the graph. Let's draw a line parallel to 0 through point 2 on axis 0. It intersected our graph at point 2;4. Let's drop a perpendicular from this point to axis 0 and... oops! What is the meaning there? Everything is correct, 4. This is what our function strives for, at →2. Difficult? Well, no, of course not! You probably noticed that if you substitute the value 2 into the function, the answer will be the same. Absolutely right. This is how these “complex” limits are solved. Don't forget to check for certainty! Certainty is when we have a clear result. Uncertainty when there is no clear result. For example: or - all this is uncertainty. This is very important, never forget about it! Therefore, you should have the following entry in your notebook (don’t forget to draw a picture): lim → lim → 2 4 Well, with this, in general, everything is clear. Practice and calculate these limits: lim → ! 1 #;lim → ;lim → ;lim → √ The same happens for the case when →∞ or to another infinite number: lim → ∞ ∞ And here is an example where there is uncertainty: lim → sin If we substitute for the value, equal to 0, then this is what we get: . And this is uncertainty, therefore, we have no right to decide! Then I will teach you how to reveal uncertainty. Now you must not forget about this. They framed and checked. Is it being decided? This means certainty. Can't decide? Well, then decide later. When you get through everything. Let's move on to formalities, that is, to definitions. Higher mathematics for dummies. Limit of a function 2011 5 UNDEFINITENESS, 0 , 1 , ∞ , 0 ∙ ∞ , ∞ ∞ Definition 1 (limit of a function according to Cauchy) No. 1. Prove that lim → sin0. For convenience, let's formulate a theorem (according to Cauchy) for our case. Here's what we get: Let's use the inequality | sin | (| | ∀. Let us set an arbitrary * 0 and set +*. Then if | | ,+, then | sin | (| | ,+*. This means (according to the definition of a function according to Cauchy) that lim → sin0. Therefore there is basically nothing to explain about this. As for | sin | (| | this just needs to be remembered. As for *, this is a very small number located in the neighborhood. No. 2. Using “* +” reasoning, prove that lim → 4. Fill in the following table: * 0.1 0.01 0.001 0.0001 … + The number b is called the limit of the function at a point (as →), if ∀ 0 ∃ 0 such that ∀ satisfying the conditions, 0 | |, the inequality || holds. The number 0 is called the limit of the function sin at point 0 (as → 0), if ∀ 0 ∃ 0 is such that ∀ satisfying the conditions 0 | | , the inequality | sin | is satisfied. Higher mathematics for dummies. Function limit 2011 6 Let * 0 be arbitrary. Then | 4 | | 2 4 2 | (| 2 | 4 | 2 | (*, as soon as 0, | 2 | , √ 4 * 2 √ . The last inequality will be even more true if * √ 4 * 2 * 2 √ 4 * * 2 √ 4 4* * * 22 * ​​+ * | 2 |. So, let's still look at this example in more detail. 1) Let's write down the definition: The number 4 is called the limit of the function at point 2 (at →2), if ∀* 0 ∃+ 0 such that ∀, satisfying the conditions 0, 0, | 2 | ,+, the inequality | 4 | ,* is satisfied. 2) Let us simplify: a) Condition: 0, | 2 | ,+ +, 2,+ 2 +,2 + b) inequality: | 4 | ,* *, 4,* 4 *,4 * 3) Let us understand: The number 4 is called the limit of a function at point 2 (as → 2) if ∀* 0 ∃+ 0 such that ∀ satisfying the conditions 0, 2 +,2 +, the inequality 4 *,4 * is satisfied. All! Read the last definition we wrote using a graph. Right? Well of course it's true! I wrote this method specifically for you to understand. You will not find this in any literature. Therefore, if you want to really solve all this quickly - go ahead! Yes, to explain how this is done analytically, I am not Higher Mathematics for Dummies. Function limit 2011 7 I'm sure I can. I wrote you an example, now you must figure it out yourself using my graphical method. Everything is built from understanding, gentlemen. Now I’ll try to explain everything on an analytical level. No. 3. To secure. Prove, using Cauchy’s definition of the limit of a function, that lim → −16 −4 = 2 Step 1: Let us define the function () , which is our expression under the limit sign: = −16 −4 Since we are considering a limit tending to 4, you need to consider some neighborhood of 4, which is defined for this function. For example, the interval is from 2 to 5. 40(2.5) But! Please note that our function is not defined everywhere! It is not defined at 0 and at = 4. I hope you understand this, but just in case I’ll write it down: −4 ≠ 0 → −4 ≠ 0 → 2 ≠ 0 ≠ 4 . I hope everything is clear. Okay, we got distracted, so let’s quickly move on. We can, in principle, consider any interval, but this one is more convenient for us than 40(2.5). Step 2: Let's write down the definition of the limit of the function () according to Cauchy. ∀* > 0,∃+ > 0:∀ ≠ 4, | −4 |< + ⇒ | −2 | < * Это значит: для любого * мы должны найти такое+, что как только x у нас отлично от 4 и x-4 по модулю не превосходит + ⇒ | −2 | должно не превосходить*. Шаг 3: Преобразуем выражение | −2 | , ≠ 4. Высшая математика для чайников. Предел функции 2011 год 8 | −2 | = 3 −16 −4 −23 = 4 +4 −2 4 = | −4 | Эти преобразования нетрудно проделать самостоятельно. Надеюсь, у вас не вызывает это трудности. Итак, ∀* > 0,∃+ > 0:∀ ≠ 4, | −4 | < + ⇒ | −2 | < * и | −2 | = | | . Заметьте, информации все больше и больше! Шаг 4: Оценим сверху выражение | −2 | , ≠ 4, ∈ (2,5). 3 −16 −4 −23 < | −4 | 2 Поняли? Мы оцениваем | | , т.к. 5 −2 5 = | | . Следовательно, | | >| | . The most important thing here is not to get confused. ∈ 2.5 – we set this condition at the beginning. This is where fractions are compared. What's more | | or | | , where ∈ 2.5. Of course the first fraction. Where the denominator is smaller, the fraction is larger (with the same numerators). Step 5: Set + = 2*. Here we can take just *, we can also take 5*. In this case, it is most convenient for us when + = 2*. So here's what we have now: ∀0 2.5 0< | −4 | < + | −2 | < + 2 = * Вывод: Все! Мы доказали, что предел равен 2. Вывод один: если хотите решать все это, берите еще раз и решайте. И так до тех пор, пока не поймете. Я попытался описать, как это доказывается аналитически. Можете посмотреть на это все и с графической точки зрения, не забыв все упростить. Информация: Вообще, честно говоря, от Вас таких доказательств не должны требовать. Они слишком уж “плавающие”. Если Вам все же интересна эта тема, откройте любой Высшая математика для чайников. Предел функции 2011 год 9 учебник и посмотрите там материал. Соответственно, Вы ничего не поймете, если не напишете собственноручно решение + графики. Это Вам небольшая подсказка. Нарисуйте! И все сразу станет ясно. №1. Я забегаю немного вперед, но хотелось бы решить этот предел: lim → 16 4 Если мы подставим 4 под, у нас получится неопределенность: lim → 16 4 7 00 8 неопределенность! Что делать? Все просто. А давайте ка упростим дробь! 16 4 4 4 4 4 Все! Теперь, если мы подставим 4, у нас будет определенность, а, следовательно, мы можем решать. lim → 16 4 lim → 4 7 84 8 2 Вывод: от неопределенности мы избавляемся с помощью преобразований. №2. Посчитать предел: lim → 4 6 16 Здесь все очень просто. Разложим на множители числитель и знаменатель. Рассказываю первый и последний раз, как это делать. Что бы разложить знаменатель на множители, мы должны приравнять его к нулю и просто решить уравнение. Давайте сделаем это. 6 160 Что бы решить квадратное уравнение, прежде всего нужно найти дискриминант по формуле: D 4E Высшая математика для чайников. Предел функции 2011 год 10 ,E − элементы квадратного уравнения. В общем виде квадратное уравнение выглядит так: + +E = 0 Следовательно, в нашем случае = 1, = 6,E = −16. Подставляем значения и находим дискриминант: D = 36 +4 ∙ 1 ∙ 16 = 100 Далее находим корни квадратного уравнения, используя формулу, = − ± √ D 2 Подставляем и получаем: , = −6 ±10 2 = F = −6 +10 2 = 2 = −6 −10 2 = −8 Корни нашли, а значит мы очень близки к разложению на множители квадратного многочлена. Сначала запишем формулу: + +E = (−)(−) Заметим, что не всякий многочлен можно так расписать. В данном случае у нас нет никаких противоречий, и, следовательно, это можно делать. Таким образом: +6 −16 = (−2)(+8) Вот эту вещь вы должны уметь делать очень быстро. Ну, максимум – минута. Так что, если есть проблемы, сразу же их решайте. В числителе можно тоже разложить на множители. Это сделать гораздо проще, так как там разность квадратов. Напоминаю формулу: − = (−)(+) Таким образом: −4 = (−2)(+2) И получаем наш предел: lim → −4 +6 −16 = lim → (−2)(+2) (−2)(+8) = lim → (− 2) (+2) (− 2) (+8) = lim → +2 +8 = 4 10 = 25 Как видите, в общем-то решение в одну строчку. №3. Посчитать предел: Высшая математика для чайников. Предел функции 2011 год 11 lim → +5 +4 2 + −1 = lim → (+1)(+4) (2 −1)(+1) = lim → (+ 1) (+4) (2 −1)(+ 1) = lim → +4 2 −1 =− 33 = −1 №4. Посчитать предел: lim → − +2 −5 +3 +4 −7 +2 Здесь я вас хочу научить одной хитрой штучке. Как разложить на множители многочлен, у которого степень > 2? We cannot do this using the discriminant - it is only for quadratic equations. So what to do? I explain: in order to factorize our numerator, we only need to find at least one root. In this case, we have no choice but to select. − +2 −5 +3 = 0 When is the equality true? After thinking a little, we answer: when = 1. Right? Substitute 1 into the equation and you will see this. Next, we have the right to factorize our polynomial: − +2 −5 +3 = (−1) ∙ G() G is the function that we have to find. We solve the equation for G(). We get: G = − +2 −5 +3 −1 Well, now we just divide one by the other in a column! − − + 2 − 5 + 3 − 1 − + 2 − 3 = G () − 2 − 5 + 3 2 − 2 − − 3 + 3 − 3 + 3 0 Thus, our function is expanded as follows: − +2 − 5 +3 = (−1) ∙ (+2 −3) We do the same with the denominator and get: +4 −7 +2 = (−1)(+5 −2) Higher mathematics for dummies. Limit of the function 2011 12 Total: lim → 2 5 3 4 7 2 lim → 1 2 3 1 5 2 lim → 2 3 5 2 1 2 3 1 5 2 04 0 No.5. Calculate the limit: lim → sin cos tg 1 lim → sin cos sin cos cos cos lim → sin cos sin cos cos lim → sin cos cos sin cos lim → cos √ 2 2 Definition 2 (the limit of a function according to Heine) The limit of a function according to Heine is rare can be found somewhere in practice. All you need to do is learn it just in case. It might be useful. We emphasize that the concept of the limit of a function at a point is introduced only for the limit points of the domain of definition of the function. Note that in this case the function may not be defined at the point, i.e., generally speaking, it does not belong. The number b is called the limit of a function at a point if for any sequence converging to! such that ∈ , # , the corresponding sequence of function values! converges to b. Designation: lim → or → when → . Higher mathematics for dummies. Limit of a function 2011 13 Definitions 1 and 2 of the limit of a function are equivalent. Let and O be defined in some neighborhood of the point, except, perhaps, the point itself, and lim → , lim → OE. Then: lim → P O Q E ; lim → P O Q E lim → O E ; lim → O E subject to E 0 Let,O and T be defined in some neighborhood of the point, except perhaps the point itself, and satisfy the inequalities (O (T. Let lim → lim → T. Then lim → O. Here, it seems, everything is clear. The theorems are expressed clearly and clearly, the information should be perceived easily. If something is wrong, don’t worry, examples are waiting for us ahead. 2. Theorems about limits Higher mathematics for dummies. Limit of a function 2011 14 One-sided limits... Not too positive sounds, doesn't it? In fact, everything is very simple. In Fig. Figure 3 shows the graph of the function. Let's try to take a couple of limits. I think we will succeed! 1) If →1. lim → 1 7 11 is certainty 8 1 2) If →0. lim → uncertainty Therefore, we have no right to decide further, and there is no way to simplify. Therefore, there is no limit. Look at fig. 3 and you will see that the function is not defined there, next. There can be no talk of any limit. 3) If →0 0. Writing →0 0 in this case means “look at how the function to the right of 0 behaves.” And what do we see on the graph? The function increases to + infinity. Therefore: lim → 1 7 1 0 0 certainty 8 ∞ Do you understand? 0 0 0, therefore, we are no longer dividing by zero. Let's look at the following examples. 4) If →0 0. What does the function to the left of 0 do? That's right, it's decreasing. Moreover, it decreases towards ∞. lim → 1 7 1 0 0 certainty 8 ∞ How do you like it? 5) If →∞ 3. One-sided limits Fig. 3 Higher mathematics for dummies. Limit of the function 2011 15 We look at the graph and see that the function tends to 0 as →∞. lim → 1 7 1 ∞ certainty 8 0 6) If →∞ Everything is the same: lim → 1 7 1 ∞ certainty 8 0 The last two I recommend remembering the example. When uncertainty is revealed, we will really need them later. Well, do you get the point? Well, then formalities... Definition 1 (limit of a function according to Cauchy) Definition 2 (limit of a function according to Heine) In general, there is nothing to add here. There is a complete analogy with the previous definitions by Cauchy and Heine, so if you understand how limits are proven, you can prove one-sided ones. The structure of the evidence is the same. Notation: lim → && 0 If 0 and 0 exist, and 0 0 , then lim → exists. The number b is called the right (left) limit of a function at point a if for any sequence converging to a! such that, the corresponding sequence of function values! converges to b. A number b is called the right (left) limit of a function at a point a if ∀ 0 ∃ 0 such that ∀ satisfying the conditions & (, the inequality | | is satisfied. Higher mathematics for dummies. Limit of a function 2011 16 If the function is defined in a certain neighborhood point a, with the exception, perhaps, of point a itself, and there is lim →, then there are 0 and 0, and 0 0. Just in case, let's consider an example for Theorem 4. Let's consider the function √. It is shown in Fig. 4. Let's find the limits: lim → √ V √ 4 0 certainty W 2 Why didn’t 0 affect anything? Yes, because he has no need to change anything. The function is defined in 4, therefore, there is no need to take 0. lim → √ V √ 4 0 definition W 2 Everything is the same. The function is defined at 4, therefore, there is no need to take 0. Nobody explains this, because it is all quite logical. Hence, by Theorem 4: lim → √ ,lim → √ exist, and lim → √ lim → √ 2 Therefore, there is a limit lim → √ 2. So, this is fixed. What if we consider 0? Well, let's check: lim → √ V √ 0 0 certainty W 0 This limit exists. Look at the function and you will see that it is defined there. lim → √ V √ 0 0 uncertainty W limit does not exist Remember once and for all: the root cannot be negative! Therefore there is no limit! But there is this: lim → √ V √ 0 certainty W 0 As you can see, Theorem 4 only works in one direction. You can't put a negative in it. Therefore, friends, be careful! Rice. 4 Higher mathematics for dummies. Limit of function 2011 17 We have already considered some cases (uncertainty disclosure (part 1)). We get rid of uncertainty with the help of transformations! Please remember this and don’t be afraid of anything. And now I want to tell you one little secret: if →∞, then in most cases the expression under the limit sign should be converted to forms of the form E ⁄, where c is a number. Why? Because this fraction will always tend to 0! You and I have already proven this. Remember and always use this! No. 1. Calculate the limit: lim → 5 lim → ]1 5 ^ lim → !1 5 # 1 0 1 How do you like it? Conclusion: when we have a fraction, we take it out → reduce it → write the answer. P.S. Now I will not write the word certainty in square brackets☺ No. 2. Calculate the limit: lim → 2 lim → 4 4 lim → ] 1 4 4 ^ lim → ! 1 4 4 # 0 0 0 0 Cool? Yes! So, let's make one more observation: in such cases we put out the same degree as in the denominator. Although, if the highest degree is in the numerator, then it is better to take it out. In general, whatever is more convenient for you. You can do it this way and that way. No. 3. Calculate the limit: lim → 4 2 ∞∞ uncertainty lim → 8 16 4 4 lim → ] 8 16 ^ ]1 4 4 ^ lim → 8 1 4 4 lim → ]1 8 ^ ] 1 1 ^ lim → 1 8 1 7 10 8 ∞ No. 4. Calculate the limit: lim → " 0 4. The limit of the function at (→ ∞ Higher mathematics for dummies. Limit of the function 2011 18 lim → 3 4 1 2 5 2 lim → ]3 4 1 ^ ] 2 5 2^ lim → 3 4 1 2 5 2 7 3 0 0 0 0 2 8 32 No. 5. Calculate the limit: lim → 2 5 1 6 1 lim → ]2 1 5 1 ^ ]6 1 1 ^ lim → 2 1 5 1 6 1 1 lim → 2 1 6 ∞ No. 6. Calculate the limit: lim → 1 2 4 4 lim → ] 1 2 4 ^ ] 4 1^ lim → 1 2 4 4 1 7 0 0 0 0 1 8 0 I repeat once again, when it’s a fraction, then we take it out ! It's time to tell you a second secret. If we are given an expression of the form _ `_ , do not be lazy to multiply it by. Here's an example: lim → ∞∞uncertainty lim → ∙ lim → 2 lim → 2 1 ]1 1 ^ lim → 2 1 1 1 lim → ]1 2 1 ^ ] 1 1 ^ 7 10 8 ∞ Undoubtedly, in the future you will not You will describe everything in detail. A few steps will be enough for you, so don't worry. P.S. As soon as you meet #1. Calculate the limit: lim → b 8 3 b Is it difficult? No! What kind does it look like? On _ `_ . Let's do the conjugate. & & CONNECTED Higher mathematics for dummies. Limit of function 2011 19 lim → b +8 +3 − b + = lim → P√ +8 +3 − √ + QP√ +8 +3 + √ + Q √ +8 +3 − √ + = lim → +8 +3 − − √ +8 +3 − √ + = lim → 7 +3 c 1 + 8 + 3 + c 1 + 1 = lim → ]7 + 3 ^ d c 1 + 8 + 3 + c 1 + 1 e = lim → 7 + 3 c 1 + 8 + 3 + c 1 + 1 That's what I told you. You should ALL end up with fractions of the form c because they all tend to 0!!! Continue: lim → 7 + 3 c 1 + 8 + 3 + c 1 + 1 = 7 7 +0 √ 1 +0 +0 + √ 1 +0 8 = 72 Scary? Well no☺. Slowly, take your time, push your limits and you will achieve great things! No. 2. Calculate the limit: lim → c + b + √ √ +1 Scary☺? Don't worry, everything is the same. Something needs to be cut. What and how? √ − this needs to be taken out and shortened. If we try to make it out, then you and I will simply get confused, and the answer will not change. Unless there may be uncertainty. That is, we take out x with the highest power in the denominator. lim → c + b + √ √ +1 = lim → √ ∙ f 1 + g 1 + c 1 √ ∙ c 1 + 1 = lim → f 1 + g 1 + c 1 c 1 + 1 = h i i i i i j f 1 + g 10 + c 1 0 c 1 + 10 k l l l l l l m = 1 The difficulty here can only be one thing: how to make √ ? I hope you can do this. No. 3. Calculate the limit: lim → P −√ −1 Q + P +√ −1 Q Higher mathematics for dummies. Function limit 2011 20 Whoever our alien is, we will still solve it. First, let's use Theorem 2 to split our limit into two limits. It will be much easier to solve this way, in the sense that you can get less confused. If you are afraid of breaking, then suffer yourself. ☺ lim → P −√ −1 Q + P +√ −1 Q = lim → P −√ −1 Q + lim → P +√ −1 Q = lim → d −√ −1 e + lim → d +√ − 1 e We just simplified everything for further work with limits using addition of fractions and the property of powers. Now we have two limits. We see a fraction. How did I teach you? That's right, we see a fraction - multiply it by its conjugate. So let's do this together. lim → d −√ −1 e + lim → d +√ −1 e = lim → d P −√ −1 QP +√ −1 Q ∙ P +√ −1 Q e + lim → d P +√ −1 QP −√ −1 Q ∙ P −√ −1 Q e This is what we got. Note that we are doing the same thing as before. The only difference is the size. Now we need to simplify each limit. In the numerator we have the difference of squares. Let's simplify the first limit: lim → d P −√ −1 QP +√ −1 Q ∙ P +√ −1 Q e = lim → n − P √ −1 Q ∙ P +√ −1 Q o = lim → d − + 1 ∙ P +√ −1 Q e = lim → d 1 ∙ P +√ −1 Q e The first one has been simplified. Now let's move on to the second: lim → d P +√ −1 QP −√ −1 Q ∙ P −√ −1 Q e = lim → d 1 ∙ P −√ −1 Q e Here's what we got: lim → P −√ −1 Q + P +√ −1 Q = lim → d 1 ∙ P +√ −1 Q e + lim → d 1 ∙ P −√ −1 Q e We see the fraction. What to do? TAKE IT OUT! First limit: Higher mathematics for dummies. Limit of the function 2011 21 lim → d 1 ∙ P +√ −1 Q e = lim → ! 1 + ∙ √ −1 # = lim → p q r ∙ 1 d 1 + c 1 − 1 e s t u = 7 02 8 = 0 Second limit: lim → d 1 ∙ P −√ −1 Q e = lim → ! 1 − ∙ √ −1 # = lim → p q r ∙ 1 d 1 − c 1 − 1 e s t u = 7 00 −uncertainty! 8 Friends, this is what you will encounter often, especially in large examples. What to do? The answer is simple: go back and do it differently. It’s good that at least we have calculated the first limit. Well, let's go back to breaking down the limits. Here's what we had: lim → d +√ −1 e How to solve if our method did not work? What to do if the “conjugate method” does not work. Let's try to take it out right away? We take it out with the highest power in the denominator, so it’s simple. lim → d +√ −1 e = lim → p q r d 1 + c 1 − 1 e s t u = lim → n1 + g 1 − 1 o = V 1 + √ 1 −0 W = 2 It turns out, in fact, everything was somewhat simpler . Total: lim → P −√ −1 Q + P +√ −1 Q = lim → d −√ −1 e + lim → d +√ −1 e = 0 +2 That’s it! Answer: 2 Difficult? I don't think so. The main thing here is accuracy and perseverance. If it doesn’t work out right away, don’t give up everything. No. 4. Calculate the limit: Higher mathematics for dummies. Limit of the function 2011 22 lim → √ 4 − − √ 4 + 3 Here we do not tend to infinity, but I want to show that the adjoint method also applies here. lim → √ 4 − − √ 4 + 3 = lim → P √ 4 − − √ 4 + QP √ 4 − + √ 4 + Q 3 P √ 4 − + √ 4 + Q = lim → 4 − −4 − 3 P √ 4 − + √ 4 + Q = lim → −2 3 P √ 4 − + √ 4 + Q = − 23 lim → 1 √ 4 − + √ 4 + = − 16 No.5. Calculate the limit: lim → √ +1 −1 √ +2 − √ 2 Here we’ll do it even cooler - multiply the numerator and denominator by the conjugate expressions of the numerator and denominator. lim → √ +1 −1 √ +2 − √ 2 = lim → P√ +1 −1 QP√ +1 +1 QP√ +2 + √ 2 Q P√ +2 − √ 2 QP√ +1 +1 QP√ +2 + √ 2 Q = lim → (+1 −1) P√ +2 + √ 2 Q (+2 −2) P√ +1 +1 Q = lim → P√ +2 + √ 2 Q P√ +1 +1 Q = lim → √ +2 + √ 2 √ +1 +1 = √ 2 No.6. Calculate the limit: lim → b 1 +tg − b 1 −tg sin2 = lim → P b 1 +tg − b 1 −tg QP b 1 +tg + b 1 −tg Q sin2 P b 1 +tg + b 1 −tg Q = lim → 2tg sin2 P b 1 +tg + b 1 −tg Q = lim → 1 cos P b 1 +tg + b 1 −tg Q = 12 Higher mathematics for dummies. Function limit 2011 23 So, what conclusion can we draw from all the previous? Well, firstly, if you are asked to calculate the limit, then for sure there is uncertainty. I recommend that you memorize the signs below!!! Example: lim → lim → lim → 2 lim → ]1 2 1 ^ ] 1 1 ^ lim → 1 2 1 1 1 7 1 0 0 0 0 8 ∞ & & , DISCLOSURE PREREQUISITES 2) If we have an expression of type, and the result is uncertainty, then we need to carry out the following operation: and then remove and reduce so that in all cases there is in denominator , DISCLOSURE OF DEFINITION 1) If we have an expression of type, and the result is uncertainty, then we need to carry out the following operation: a then take it out and reduce it so that in all cases it is in the denominator. Higher mathematics for dummies. Limit of the function 2011 24 Example: lim → lim → lim → lim → ]1 1 ^ ] 1 1 ^ lim → 1 1 1 1 7 1 0 0 0 8 ∞ As you can see, we calculated the same limit in different ways. This doesn't always happen! You must memorize all the tables like a multiplication table. Probably, many may have a question: when to use what? Practice, friends. You have no other choice, and cannot have it. Only through your own experience can you achieve some results. As always, let's move on to the formalities (professor's theory):) * "*+ , D R A S C R Y T I N E O D E R D E N I N S 3) If we have an expression like Then you need to either immediately take out and reduce so that in all cases it is in the denominator, or multiply by the conjugate of the numerator or denominator. Depending on the situation. You should use all three of the above points when disclosing uncertainty when → ∞. If tends to some other value, and we have uncertainty, then we simply use simplifications (conjugate or abbreviations) Let the function be defined on the line ", & ∞. A number is called the limit of a function as → & ∞ lim → if ∀ 0 ∃ , 0 - " such that ∀ , the inequality | | is satisfied. Higher mathematics for dummies. Limit of a function 2011 25 Let the function be defined on the line " , & ∞ . Number is called the limit of a function at → & ∞ if for any infinitely large sequence! " the corresponding sequence of function values! converges to. Higher mathematics for dummies. Limit of a function 2011 26 The same is true for infinitesimal functions. In my opinion, we need a definition either for evidence or... for other purposes. At least I never needed it. So, we have already seen examples before where the limit was equal to ∞. As you can see, they are counted exactly the same as all the others. The key role here is played by this construction: V 1 0 v W . Remember, this construction is ALWAYS equal to ∞! | | . . A function is said to be infinitely large at a point a on the right if ∀ . 0 ∃ 0 such that ∀ satisfying the condition &, the inequality is satisfied. Notation: lim → ∞ A function is said to be infinitely large for → & ∞ if ∀ . 0 ∃ , - " such that ∀ , | | . . Notation: lim → ∞ 5. Infinitely large functions 0 1 0 1 2 ∞ Higher mathematics for dummies. Function limit 2011 27 Yes, this is exactly what we have to do now. They we will REALLY need them in the future. Therefore, it is important to consolidate them now, and at the same time calculate the limits. I agree, this is tedious and uninteresting. If you know something, skip and move on, I allow☺. So, this is our first and the most important function. We already covered it earlier, but let's repeat what we've already done. lim → w ∞ lim → w 0 lim → w ∞ lim → w 0 If you want, you can memorize all this, but in general, I recommend that you remember the graph itself. In my opinion, everything is quite clear. Well, you simply must know this function, but, just in case, I will remind you of it. You know, there are different cases☺. lim → ∞ lim → ∞ 6. Graphs of elementary functions 3 1 & & "Higher mathematics for dummies. Limit of the function 2011 28 The function has its own name - an exponential function. Here it is important not to forget about one thing: at 1 the function increases; at 0.1 the function decreases. Here let's look at examples: #1. Calculate the limit 1 lim → 2 2 ∞ lim → 2 2 0 MEMORY! This is something you simply must memorize, because graphs are often confused with each other. No. 2. Calculate the limit 0.1 lim → ! 12 # lim → 1 2 7 1 2 1 ∞ 8 0 lim → ! 12 # lim → 1 2 7 1 2 10 8 ∞ As you can see, we simply derived the last two limits from the previous two. Memorize! The function has its own name - logarithmic function. There are also two pitfalls here: at 1 the function increases; at 0.1 the function decreases. No. 1. Calculate the limits 1 lim → log 0 lim → log ∞ lim → log ∄ lim → log ∄ №2. Calculate limits 0.1 log Higher mathematics for dummies. Limit of a function 2011 29 lim → log 0 lim → log ∞ lim → log ∄ lim → log ∄ I’m sure you won’t remember so much, so it’s better to memorize the graph. Ok! Let's move on... The function has its own name - sine wave. No. 1. Calculate the limit lim → sin. What to do? The graph clearly shows that the function “jumps” from one value to another. Conclusion: there is no such limit. Let's just look at examples where the function tends to different values: lim → sin ( | ) | ~ lim → sin1 lim → sin 0 lim → sin 1 ; I'll do the same for the cosine wave. No. 1. Calculate the limit: lim → cos. All the same thoughts. There is no limit! This is what we get: lim → cos ( | ) | ~ lim → cos0 lim → cos 1 lim → cos 1 ; sin "67 Higher mathematics for dummies. Limit of a function 2011 30 The figure shows two functions: O and EO. As you can see, they are very similar, so it is very important whether you remember them or not. Let's do a little experiment. Try to remember the two graphs. Once you are sure that you have learned everything, solve all the limits below, and then check yourself on the graphs. No. 1. Calculate the limits: lim → tg lim → tg lim → tg lim → tg lim → tg lim → tg lim → ctg lim → ctg lim → ctg lim → ctg lim → ctg lim → ctg arcsin – inverse function to the sin function. arccos – inverse function to the cos function. No. 1. Calculate the limit: lim → arcsin. Let's look at the arcsin graph. What do we see ? At → 0, the function takes on infinitely many values. For example, lim → arcsin0 and lim → arcsin, etc. We conclude: our graph has a period. lim → arcsinw,w is an integer lying in the interval ∞,∞ 89 "89 arcsin arccos Highest mathematics for dummies. Function limit 2011 31 Same with arccos. arctg is the inverse function to the tg function. arcctg is the inverse function to the ctg function. No. 1. Calculate the limit: lim → arctgw ∙ 2 w an integer with a step of 2. I.e. lim → arctan ⋯. You can write it like this: lim → arctan 2 2 2 w Note that this is an arbitrary integer that we set ourselves. This concludes our section – graphs of elementary functions. From the author: Congratulations! You were able to complete the first chapter “The Limit of a Function” of the first part “The Limit and Continuity of a Function”. Of course, that's not all. I told you only basic things. Next we will have the first wonderful and second wonderful chapels and other methods of taking limits. If you understand everything I wrote here, then it will only be interesting! Nothing super complicated awaits you... arctg arcctg Higher mathematics for dummies. Limit of a function 2011 32 Chapter 2. Continuity of a function at a point. Remember this definition once and for all! If you don't know it, you are nothing and no one in mathematics. Let's look at a simple example: 1 Task: check the function for continuity at points 1;0. 1. 1. Using definition 1, we get: lim → 1 1 ↭ 1 11 1 Does definition 1 hold? Yes! lim → 1 1 1 Conclusion: the function is continuous at the point 1. 2. 0. Using definition 1, we get: lim → 1 ∞↭ 0 10 →∄ Does definition 1 hold? No! lim → 1 0 lim → A function is called continuous at point a if 1. Continuity of the function at the point. Contents: 1) Continuity of a function at a point 2) Continuity of a complex function 3) Classification of discontinuity points 4) Continuity of elementary functions 5) The first wonderful limit 6) The second wonderful limit 7) Briefly about Maple Higher mathematics for dummies. Limit of a function 2011 33 Conclusion: the function does not exist at point 0. The same thing here. Please look into functions such as ln and others for yourself. Although, I think that everything is extremely clear. In order for a function to be continuous at, it is necessary and sufficient that it be continuous at this point on the right and on the left. If the functions and O are continuous at a point, then the functions O, O, O, /O are also continuous at the point (quotient - under the condition O 0). Example No. 1. Examine the continuity of a function. To begin with, let us describe the domain of definition D∞,0 ∪0,∞, since the denominator cannot be equal to 0. Now we simply use Theorem 6: lim → , where 0. Therefore, by Theorem 6, the function is continuous at any point except 0. lim → > respectively lim → E . Let the function be defined in the right (left) semi-neighborhood of the point a, i.e. on some half-interval, & (respectively,). The function is said to be continuous on the right (respectively on the left) at point a, if Higher Mathematics for Dummies. Function limit 2011 34 However, you won’t really need this for now. Here are examples of complex functions: b | sin | ,cos 1 ,log 1 . Why are they complicated? Let's look at a chain of sequential transformations for the first of them: sin | | √ . That's all! Now let's move on to the second function: 1 cos. And so on. I don't want to spend a lot of time on this. I hope you understand everything already. Well, let's move on to the theorem. Let a function be continuous at a point, and a function be continuous at a point. Then the complex function P Q is continuous at the point. Let's look at an example for evidence. This is where we need to consider a complex function. Example No. 1 Prove that: lim → 1 ln, 0, 1. Consider function 1. It is continuous at the point 0 and 0 0. Moreover, Let the function F be defined on a set, and G the set of values ​​of this function. Let, further, let a function H be defined on the set G. Then they say that a complex function is defined on the set, and write H, where F, or HF. 2. Continuity of a complex function. Higher mathematics for dummies. Function limit 2011 35 log 1, 1 log 1. Let's calculate lim → : lim → log 1 lim → ln ln 1 This step may not be clear, so I must remind you of the formula for converting to a logarithm with a different base: Remember it and don't come back to it again. In this case, a new foundation. Let's write a formula specifically for our case: log 1 log 1 log ln1 ln. So, we continue: lim → log 1 lim → ln ln 1 ln 1 lim → ln1. Right? ln is a number, so we took it out. Now we need to calculate the limit lim → . Let's represent the function in the form ln 1 ln (also a property of the logarithm!), where 1 . Since lim → 1 (This is the second wonderful limit. We haven’t passed it yet, but believe me, the equality is true), and the function ln is continuous at a point, then lim → ln 1 ln1. Let's return to our example. And this is what we get: log log log ∙ log log Higher mathematics for dummies. Limit of the function 2011 36 lim → log (1 +) = lim → ln ln 1 + = ln 1 lim → ln(1 +) = ln 1 = ln. Let us now consider the function (), continuous at the point = 0: = log (1 +) for ≠ 0 lnat = 0 According to Theorem 8, the complex function P Q = −1 at ≠ 0 lnat = 0 Is continuous at the point = 0. Therefore lim → − 1 = ln. Difficult? Maybe, but you have to look into it because it is very important to understanding this topic. Moreover, it requires attentiveness and “a little thinking.” Higher mathematics for dummies. Function limit 2011 37 First, let's understand what a “breaking point” actually means. Everything is extremely simple! Before you start considering the classification of discontinuity points, you should always check the condition: must be defined in some neighborhood of the point, with the possible exception of the point itself. If the condition is met, then the classification of discontinuity points can be considered. Example No. 1. sin First of all, let's write the domain of definition: D ∞;0 ∪0;∞. From here it is immediately clear that 0 is an unusual point. In it, the function is not defined, but is defined in its neighborhood. lim → sin 1 0 sin . It follows that 0 is a removable discontinuity point. A point is called a discontinuity point of a function if it is not continuous at this point. lim → # Point – removable break point if 3. Classification of break points. Higher mathematics for dummies. Function limit 2011 38 Example No. 1. sgn The sgn function should already be known to you before, but I will remind you of it. sgn 1,0, 1, 0 0 ,0 , lim → sgn 1, lim → sgn 1, 0 0. It follows that lim → sgn lim → sgn sgn point 0 point of discontinuity of the first kind. Example No. 1. tg First of all, let's write the domain of definition D \ 2 w ,w0. lim → tg∞ ∃ lim → # lim → # A point is a discontinuity point of the first kind, if a Point is a discontinuity point of the second kind, if at least one of the one-sided limits does not exist or is equal to infinity. f(x) = sgn(x) Higher mathematics for dummies. Limit of the function 2011 39 lim → tg∞ Because at least one of the limits is equal to infinity, then w is a discontinuity point of the second kind. Example No. 2. ln First of all, let's write the domain of definition D 0;∞. limln → 0 limln → ∄ Because at least one of the limits does not exist, then 0 is a discontinuity point of the second kind. So, we now know the classification of breakpoints. We have looked at examples for each case. They are quite easy, so let's practice some more. In all the following numbers, determine the break points. P.S. First, try to do it yourself, and then test yourself. Good luck ☺! No. 1. 2 , ln, (1 1 lim → lim → ln0, lim → lim → 1. lim → lim → In point 1, the function has a discontinuity of the first kind. No. 2. First of all, we write: D ∞,0 ∪0,∞. Higher mathematics for dummies. Limit of a function 2011 40 lim → lim → 7 0 8 0, lim → lim → ∄. 0 limit point of the second kind. No. 3. 1 2 3 First of all, we write: 4 0 D ∞,4 ∪4 ,∞.lim → 1 2 3 lim → 1 2 3 7 1 2 0 8 12 , lim → 1 2 3 lim → 1 2 3 7 1 ∞ 8 0. 4 discontinuity point of the first kind. No. 4. | 1 | First of all , let's write. We define critical points like this: 0 1 0. Critical points: 0 and 1. Now let's write the domain of definition D ∞,0 ∪ 0.1 ∪1,∞. lim → | 1 | 7 10 8 ∞ 0 point of discontinuity of the second lim → | 1 | lim → 1 lim → 1 1 lim → 1 1 Higher mathematics for dummies. Limit of a function 2011 41 lim → | 1 | lim → 1 lim → 1 1 lim → 1 1 1 discontinuity point of the first kind. 0 discontinuity point of the second kind, 1 discontinuity point of the first kind. No. 5. 1 1 First of all, we write: D ∞,1 ∪1,∞. lim → 1 1 lim → 1 1 1 lim → 1 1 13 Removable discontinuity point: F 1 1 , 1 13 ,1 It is continuous at the discontinuity point and at D. No. 6. 1 1 1 1 1 1 To find critical points, you need to simplify the function. 1 1 1 1 1 1 1 1 Points: 0;1;1. lim → 1 repairable gap. lim → ∞second-city gap. lim → 0 removable gap. Higher mathematics for dummies. Function limit 2011 42 No. 7. cos cos 1 and we get: 2 2w 1 removable breakpoints. 0 point rupture of the city. I think there are enough examples. If you decide all this yourself, then you will know the topic 100%. Well, I hope this wasn't too boring. At least you won’t find so many analyzed examples anywhere. Higher mathematics for dummies. Function limit 2011 43 We have already discussed this topic in Chapter 1, paragraph 6. There we looked at graphs of elementary functions and calculated limits. Now let's move on to formalities and “professorial theory.” As you will notice, this “theory” is present in my book. For what? It's simple - I want you to not only take the chewed food, but also try to chew it yourself. If I remove this “theory”, then my work will go down the drain. Of course, you will be able to solve something, but you will not understand what and how. Therefore, I ask you to learn the theory! You will definitely need it in the near future. Well, that was a lyrical digression ☺. Let's move on to a little theory. Any elementary function defined in a neighborhood of a certain point is continuous at that point. This is where the “professor’s theory” ends, and we move on to remarkable limits. Functions I "6J78, log 0, # 1, sin, cos, tg, ctg, arcsin, arccos, arctg, arcctg are called the simplest (or basic) elementary functions. The set of all elementary functions is called a class of elementary functions. A function is called elementary if it can be obtained using a finite number of arithmetic operations and superpositions over the simplest elementary functions. 4. Continuity of elementary functions Higher mathematics for dummies. Limit of a function 2011 44 ​​A very important topic! In it we will learn to look for limits. You should get your hands on this, and I have a request to you: before looking at the solution, try to achieve something yourself. Memorize it once and for all! And never forget this formula! I’m not going to prove it, if you want, look on the Internet, it’s there for sure yes. Well, let's move on to the examples. No. 1. lim → sin. Solution: sin 1 sin, Hurray! A wonderful limit has appeared below. lim → sin lim → 1 sin 7 11 8 1. Easy? Absolutely... No. 2. lim → arcsin. Solution: Let's make a change of variable: let arcsin. Then sin and base →0 goes to base →0 (just substitute →0 under arcsin). In fact, it’s easier to write it like this: lim → arcsin 7 arcsin ↭sin → 0 ↭ →0 8 lim → sin 7 11 8 1. 5.The first wonderful limit lim → sin 1 Higher mathematics for dummies. Function limit 2011 45 Remember this method of changing a variable. It can be very useful to you in the future. No. 3. lim → arcsin. Solution: lim → arcsin lim → 1 arcsin 7 arcsin ↭sin →0 ↭ →0 8 1 lim → sin 7 11 8 1. No. 4. lim → sin2 sin3 . Solution: Transform the function as follows: lim → sin2 sin3 lim → ! sin2 2 ∙ 3 ​​sin3 ∙ 23 #. Let's take the constant factor beyond the limit sign and apply the theorem on the limit of products: lim → sin2 sin3 lim → ! sin2 2 ∙ 3 ​​sin3 ∙ 23 # 23 ∙ lim → sin2 2 ∙ lim → 3 sin3 We make the replacement, as in the previous example: lim → sin2 sin3 lim → ! sin2 2 ∙ 3 ​​sin3 ∙ 23 # 23 ∙ lim → sin2 2 ∙ lim → 3 sin3 ! 2 ↭sin2sin →0 ↭ →0 4 3 ↭sin3sin →0 ↭ → 0 # 23 ∙ lim !→ sin ∙ lim "→ 1 sin 23 ∙ 1 ∙ 1 23. No. 5. lim → sin 4. Multiply and divide the banner atelier on 4 and bring the expression under the limit sign to the first remarkable limit: lim → sin 4 lim → sin 4 4 ∙ 4 14 ∙ lim → sin 4 4 d 4 ↭4 →0 ↭ →0 e 14 ∙ lim !→ sin 14. Higher mathematics for dummies. Limit of a function 2011 46 No. 6. lim → 2tg 2. Let's represent the tangent through sine and cosine and use theorems on limits. lim → 2tg 2 lim → 2 ∙ sin 2 cos 2 lim → 2sin 2 cos 2 2 lim → sin 2 4] 2 ^ ∙ lim → 1 cos 2 d 2 ↭2 → 0 ↭ → 0 e 12 lim → sin ∙ lim → 1 cos 2 7 12 ∙ 1 ∙ 11 8 1. See, it’s a little more complicated here, but in principle, everything is the same. If you have learned elementary functions, then this should not seem difficult to you. No. 7. lim → 1 cos 2 tan. Using the double angle formulas we have: lim → 1 cos 2 tg lim → 1 cos sin tan lim → cos sin cos sin tg lim → cos sin cos sin tg lim → 2sin tg lim → 2sin cos sin cos 2 lim → sin lim → cos 2 ∙ 1 ∙ 1 2. Gentlemen, let’s learn trigonometric formulas! You'll still need them. Higher mathematics for dummies. Function limit 2011 47 There are many formulas, but it is advisable to learn them all. No. 8. lim → 8sin 4 . Multiply and divide the numerator by 4 cubed: sin K *L sinKcos L *cosKsinL cos K *L cos Kcos L ∓sinKsinL t9 K *L t9K *t9L 1 ∓t9Kt9L ct 9 K * L ct 9 K" t 9L ∓ 1 ct 9L * ct 9 K sin K &cos K 1 tg K &1 1 cos K ctg K &1 1 sin K sin2K 2sinKcos K cos 2K cos K sin K 2cos K 1 1 2sin K tg 2K 2tgK 1 tg K ctg2K ctg K 1 2ctgK sinK * sinL 2sin K *L 2 cos K ∓L 2 cos K &cosL 2cos K &L 2 cos K L 2 cos K cos L 2 sin K & L 2 sin K L 2 Higher mathematics for dummies: The limit of a function 2011 48 lim → 8sin 4 lim → 4 ] 4 ^ 8sin 4 8 lim → ] 4 ^ sin 4 d 4 ↭4 →0 ↭ →0 e 8lim !→ sin 8 ∙ 18. No. 9. lim → sin 2 4 1. In the denominator we can square the difference, and then, as always, go to a new variable. Then the limit will tend to 0, and therefore we can apply the first wonderful limit. lim → sin 2 4 1 lim → sin 2 2 ] 2 ↭ 2 →2 ↭ →2 20 ^lim !→ sin 1 1. No. 10. lim → sin3 sin4 6. Based on one of the theorems about limits, we can divide this limit into two limits: lim → sin3 sin4 6 lim → sin3 6 lim → sin4 6 12 lim → sin3 3 23 lim → sin4 4 3 ↭ 3 →0 ↭ →0 4 ↭ 4 → 0 ↭ →0 ¡ 12 lim → sin 23 lim → sin 76 . No. 11. lim → cos cos 3 . We transform the numerator using the formulas for the difference between the cosines of two angles and the sine of a double angle: cos cos32sin2 sin4sin cos, then lim → cos cos3 4lim → sin cos 4lim → cos4. Higher mathematics for dummies. Limit of a function 2011 49 The second remarkable limit is called the limit of the form. We are not going to prove this either. Perhaps someday I will write a separate book about all the proofs, but for now let’s not waste time on this and go straight to the examples. As soon as you see a bracket in a power, then first of all try to reduce it to the second limit. Let's look at the first numbers in great detail. No. 1. Calculate the limit: lim → ! 4 # We see a bracket to the power of 5, so we try to reduce it to the second remarkable limit. First, let's reduce what's inside to form 1: lim → ! 4 # lim → !1 4 # Now you need to “play” with the degree. Those. we need a view like /4. Why? The formula lim → !1 1 # could be written as lim → !1 1 # . In this case, instead of one we have four. So, this is what we get: lim → ! 4 # lim → !1 4 # lim → ¢ !1 4 # £ . In order to completely reduce this limit to our formula, we denote it by 4. Then we get: lim → 1 1 lim → 1 6. The second remarkable limit Higher mathematics for dummies. Function limit 2011 50 lim → ! +4 # = lim → !1 + 4 # = lim → ¢ !1 + 4 # £ = ¤ = 4 ↭ = 4 →∞↭ → ∞ ¥ = lim !→ ¦!1 + 1 # ! § = . As you can see, there is nothing complicated here. The algorithm of work is very simple: reducing the fraction to the form 1 + # reducing the degree to the form # ∙ ¨ replacing the variable and then simply counting according to the formula. If you're confused, don't worry. We still have time to look at a lot of examples ☺. No. 2. Find the limit: lim → ! +2 +1 # We act in the same way as last time: lim → ! +2 +1 # = lim → ! +1 +1 +1 # = lim → !1 + 1 +1 # Here we will highlight the degree after changing the variable. In this case, it is easier than trying to reduce to the second limit before replacement. This will not affect the result in any way. lim → ! +2 +1 # = lim → ! +1 +1 +1 # = lim → !1 + 1 +1 # = 2 = −1 ↭ = +1 →∞↭ →∞ = lim !→ !1 + 1 # ! = lim !→ ¦!1 + 1 # ! § lim !→ !1 + 1 # = ∙ 1 = . As you can see, there is nothing supernatural here. From here you can write a solution algorithm similar to the previous one. Reducing the fraction to the form 1 + # replacing the variable, reducing the degree to the form # ∙ ¨ and then we simply calculate according to the formula. No. 3. Find the limit: lim → d +5 +2 e Select the whole part in brackets: lim → d +5 +2 e = lim → d +2 +3 +2 e = lim → !1 + 3 +2 # = +2 = 3 ↭ = +2 3 →∞↭ →∞ = lim ! → !1 + 1 # ! = lim !→ ¦!1 + 1 # ! § lim !→ !1 + 1 # = ∙ 7 1 + 10 8 = . Higher mathematics for dummies. Function limit 2011 51 The example is completely similar to the previous one. If you understand how “it works,” then you are great and can safely move on. The big advantage here is that it is enough to know only a few methods to solve a particular limit. No. 4. Calculate the limit: lim → ! 1 2 # Select the whole part in brackets: lim → ! 1 2 # lim → 7 !1 1 # 12 8 lim → !1 1 # lim → ! 12 # 1 ↭ 1 → ∞↭ →0 lim !→ 7 1 ! 8 lim → 2 2 lim → 2 8 ∞ Next, I don’t want to look at each example in such detail, otherwise each solution will take up more than half the page. The main thing is that you understand the general idea and strive for the ideal solution, i.e. short I’ll give you one more piece of advice: first try to decide something yourself, and then check whether you did it right or not. No. 5. Calculate the limit: lim → !1 1 # Solution: lim → !1 1 # 1 ª« ªª lim → !1 1 # ∙!1 1 # ∙ lim → !1 1 # ∙ 1 No.6. Calculate the limit: lim → 1 Solution: lim → 1 1 ª« ªª lim → ! 1##7. Calculate the limit: lim → !1 2 # Higher mathematics for dummies. Limit of the function 2011 52 Solution: lim →!1 2 # 1 ª« ªª lim → n!1 2 # o No.8. Calculate the limit: lim → !1 4 # Solution: lim → !1 4 # 1 ª« ªª lim → n!1 4 # o No.9. Calculate the limit: lim → ! 3 1 # Solution: lim → ! 3 1 # 1 ª« ªª lim → ! 1 4 1 # lim → !1 4 1 # ∙ lim → №10. Calculate the limit: lim → 4 ln 2 3 ln5 3 Solution: lim → 4 ln 2 3 ln5 3 lim → 4 ln 2 3 5 3 lim → ln! 2 3 5 3 # lim → ln!1 3 5 3 # % lim → ln1. No. 11. Calculate limits: Higher mathematics for dummies. Limit of the function 2011 53 lim → d +1 +3 e I must say, this example is a little more interesting than the previous ones. Solution: lim → d +1 +3 e = lim → d 1 + +1 +3 −1 e = lim → d 1 + +1 − −3 +3 e = lim → !1 + −2 +3 # = lim → !1 + −2 +3 # ∙ ∙ = lim → ¢ !1 + −2 +3 # £ = &" → = = 1 With this I propose to finish the second wonderful limit. Further, at the end of the book you can find a lot of tasks on this topic. Of course, the answers will be attached. Higher mathematics for dummies. Limit of a function 2011 54 I would also like to make a note about the electronic calculation of limits. There is such a program - Maple, and there the limits are calculated simply with a bang. As you can see , on the left, in the window there are formula templates. Just click on them and fill in the data. Press Enter and get the answer. In the screenshot, for example, our last limit was calculated. Why do you need this program? For checks. Calculated the limit on paper, got the answer. We entered the formula into the program and checked it. It's actually a very convenient thing. From the author: Congratulations! You were able to complete the second chapter “Continuity of a function at a point” of the first part “Limit and continuity of a function”. Ahead of you is a comparison of infinitesimal functions, the symbol “Ο small” and its properties, calculating the limits of functions using asymptotic formulas and calculating the limits of exponential functions. The topics will be very important, so not only “technical” examples will be considered, but also examples and evidence. On this note, I want to wish you success! See you soon! Sincerely yours, Viosagmir I.A. 7. Briefly about Maple Higher mathematics for dummies. Limit of a function 2011 55 Chapter 3. Infinitesimal functions. A function is called infinitesimal at → (at a point) if lim → -0. Let - and ® be two infinitesimal functions as →. The functions - and® are called: a. Infinitesimals of the same order as → (at a point), if lim → - ® E 0; b. Equivalent infinitesimals at → (at a point), if lim → - ® 1 notation: -~®at → . If lim → () 0, then they say that - is an infinitesimal of a higher order at → (at the point) than ®, and write -²® at → (- is equal to “² small” from ® at →). For example, ² at →0. Similar definitions hold for the cases → 0, → 0, → ∞. It should be borne in mind that equalities containing the symbol “² small” are conditional. For example, the equality ² at →0 is true, but ² is false, since the symbol ² does not denote any specific function, but any function that is infinitesimal at →0 of a higher order than. There are infinitely many such functions, in particular, any function * (where ³ 1) is ² as →0. Thus, the equality ² for →0 means that the function belongs to the set of infinitesimal functions of higher order for →0 than. Therefore, “in the opposite direction” this equality is false: the entire set of functions cannot be reduced to one function. Nothing is clear ☺? Don't worry, we'll look at everything further with examples. But theory is needed in any case, otherwise my book ceases to be mathematical, and it becomes unclear what it is. 1. Comparison of infinitesimal functions. A function K is said to be infinitesimal as → (at a point) if lim → K 0 . Contents: 1) Comparison of infinitesimal functions 2) Properties of the symbol “o small” 3) Comparison of infinitesimal functions Higher mathematics for dummies. Limit of function 2011 56 Let's consider several examples relevant to this topic. No. 1. Is the equality 2² at →0 true? Solution: 2 ² – correct, since lim → 2 0. As you can see, the solution is in one line. Let's look at it in more detail ☺. Let's remember our definition! If lim → () 0, then they say that - is an infinitesimal of a higher order at → (at the point) than ®, and write -²® at → (- is equal to “² small” from ® at →). In our case, we denote by - 2. Next, we need to “dig up” ® from somewhere. Let's look in the definition at the words they write -²®. It follows from this that ®, judging by our example, is 2 ². Next we simply follow the definition, i.e. we write down the limit and check whether it is equal to zero or not. lim → - ® lim → 2 lim → 20 The limit is zero, therefore - 2 is an infinitesimal of a higher order at →0 (at point 0) than ®, and write 2 ²® at →. We will also construct our function graphs for clarity. The red graph is our “main” function - 2, and the green graph is the ® function. The picture shows that closer to zero, function - 2 tends to it faster than ®. All! We have analyzed this example in great detail. Further, all the examples will be identical, so I will not write the solution in such detail. Higher mathematics for dummies. Limit of the function 2011 57 In all other cases, the red graph is the function - , and the green graph is ® . No. 2. Is the equality 3² true as → 0? Solution: First, let's write down the functions - and ®. This is what we get: - 3,® Now look at the limit: lim → - ® lim → 3 3 0 The limit is not equal to zero, therefore the equality 3² is incorrect. But! Since the limit is equal to a constant, the functions 3 and infinitesimals are of the same order at point 0. No. 3. Is the equality b | | ² at →0? Solution: First, let's write down the functions - and ®. This is what we get: - b | | ,® Now look at the limit: lim → - ® lim → b | | lim → b | | b | | ∙ b | | 7 10 8 ∞ 0 The limit is not equal to zero, hence the equality b | | - wrong. Higher mathematics for dummies. Function limit 2011 58 No. 4. Is the equality true | | ² at →0? Solution: First, let's write down the functions - and ®. This is what we get: - ln | | ,® Now look at the limit: lim → - ® lim → n ln | | olim → 1 ln | | 0 The limit is zero, hence the equality | | - that's right. No. 5. Is the equality 1 cos² at →0 true? Solution: First, let's write down the functions - and ®. Here's what we get: - 1 cos ,® Now look at the limit: lim → - ® lim → 1 cos lim → 2sin ] 2 ^ lim → n sin] 2 ^ 2 o 2 1 ∙ 0 0 The limit is zero, hence the equality 1 cos² is correct. P.S. Solving such limits is already Higher Mathematics for Dummies. Function limit 2011 59 should not be difficult. If you feel like you can't cope, it's better to go back to chapters 1 and 2 and repeat everything. We already had all the limits of these types. This, as they say, is a base without which you can’t go anywhere. Since the examples are all identical to each other, first solve them yourself, and then look at the solution. If you don't do this, you won't learn anything!!! No. 6. Is the equality sin² true as →0? Solution: First, let's write down the functions - and ®. This is what we get: - sin ,® Now look at the limit: lim → - ® lim → sin lim → ! sin # 1 1 The limit is not equal to zero, therefore the equality sin ² is false. But! Since the limit is equal to unity, the functions sin and equivalent are infinitesimal at point 0. No. 7. Is the equality ² true for →0? Solution: First, let's write down the functions - and ®. This is what we get: - ,® Now look at the limit: lim → - ® lim → 0 The limit is zero, therefore equality ² is true. Higher mathematics for dummies. Function limit 2011 60 No. 8. Is the equality 1 cos² at →0 true? Solution: First, let's write down the functions - and ®. This is what we get: - 1 cos,® Now look at the limit: lim → - ® lim → 1 cos 12 The limit is not equal to zero, therefore the equality 1 cos² is incorrect. But! Since the limit is equal to a constant, the functions 1 cos and infinitesimals are of the same order at point 0. Higher mathematics for dummies. Limit of a function 2011 61 Let - and - be two arbitrary infinitesimal functions for → such that - ²® and - ²®. Then - - ²® as →. This theorem can be written as follows: ² ® ² ® ² ® . Let us formulate, along with the above, a number of properties of the symbol “² small” (everywhere we mean that - →0 and ® →0 as →). 1. ² ® ² ® ² ® 2. ² ® ² ® ² ® 3. ² E® ² ® ∀E 0 4. E² ® ² ® ∀E 0 5. ² ® ² P ® Q , ´ 2 ∈ µ ​​,w1 ,2,…, 1 6. P ² ® Q ² ® ∀ ∈ µ 7. ® ² ® ² ® ∀ ∈ µ 8. +)) ² ® , ´ 2 ∈ µ ​​Let us denote any infinitesimal as → function by the symbol ² 1 . Then property 8 will also be valid for 1: +)) ²1. 9. o P ∑ c , β , Q o β ,where c , numbers 10. ² P ² ® Q ² ® 11. ² P ® ² ® Q ² ® 12. -®² - ,-®² ® 13. If ~ ®, then - ®²- and - ®²® On this note, theory ends and practice begins. I recommend learning all the properties. They will be very useful to us in the future. The first task will be discussed in great detail. You will have to do the following tasks yourself in order to “get into” this topic. No. 1. Using the limit lim -→ .&- - 1, represent the function sinx in the form ¹ ² P Q at →0, where w1 or w2; and some numbers. 2.Properties of the symbol “O small”. Higher mathematics for dummies. Limit of a function 2011 62 Solution: First we prove that if - and ® are infinitesimals of the same order as →, i.e. lim → () E 0, then - с® ²® at →. In fact, since lim → - ® E → lim → d - ® E e 0 → lim → - E® ® 0, Then, by the definition of the symbol ²®, we have - E® ²®, or - E® ² ® for → . Using this equality, we obtain sinx ² as → 0. The last formula is called the asymptotic formula of the function sin as →0. The last term on the right side of this formula is called the remainder term of the asymptotic formula. Further, in subsequent examples, we will not prove the same thing and will proceed from what has already been proven, i.e. - E® ² ® at →. Therefore, I recommend reading the proof again, and most importantly, understanding it. No. 2. Using the limit lim -→ /. - represent the function sinx in the form ¹ ² P Q at →0, where w1 or w2; and some numbers. Solution: We use the formula - E® ² ® at → and get: cos 1 12 ² at →0. The last formula is called the asymptotic formula of the function cos as → 0. The last term on the right side of this formula is called the remainder term of the asymptotic formula. No. 3. Using the limit lim -→ - 1, represent the function sinx in the form ¹ ² P Q at → 0, where w1 or w2; and some numbers. Solution: Higher mathematics for dummies. Limit of the function 2011 63 We use the formula - E® ² ® as → and obtain: ln1 ² as → 0. The last formula is called the asymptotic formula of the function ln1 as →0. The last term on the right side of this formula is called the remainder term of the asymptotic formula. No. 4. Using the limit lim -→ √ - represent the function sinx in the form ¹ ² P Q at →0, where w1 or w2; and some numbers. Solution: We use the formula - E® ² ® at → and get: √ 1 1 1 ² at →0. The last formula is called the asymptotic formula of the function √ 1 as → 0. The last term on the right side of this formula ² is called the remainder term of the asymptotic formula. I think this will be enough for you. At an institute or college, almost no time is devoted to this. This time I wanted you to understand where this “² small” comes from, and how the asymptotic formulas are derived. As they say, a little theory will not hurt you and, of course, it is advisable to understand what comes from where. Higher mathematics for dummies. Limit of the function 2011 64 Previously, asymptotic formulas for the simplest elementary functions at →0 had already been obtained. Let's write these formulas in the form of a table. The indicated formulas remain valid if instead of the argument we substitute in them, where º » an infinitesimal sequence, or where lim → 0. For example, the representation following from the first formula is valid: sin 1 1 ²! 1 #, where 2 ² ] ^ is an infinitesimal sequence of higher order than 2, i.e. lim → ²] 1 ^ 1 lim → ²! 10. That is, by this we want to say that if 2 sin →0, then we can apply the asymptotic formula to the sine. For example, function 1 is infinitesimal as → 1, so from the third formula we obtain the equality ln P 1 Q ² as →1, or ln 1 1 1² as → 1. Here is another example. Using the previous equality and the second formula, we write the asymptotic representation of the function cos ln as →1. 1 sin & 6 2 cos 1 2 & 6 3ln 1 & &6 4 1 & ln & 6 0 5 S 1 & & 6 6 1 & 1 & & 6 7 tg & 6 8sh &6 9 ch 1 & 2 & 6 10 th & 6 Higher mathematics for dummies. Limit of the function 2011 65 The function ln tends to zero as →1, therefore it is infinitesimal, therefore we can apply asymptotic formula number three: coslncos 1 ² 1. The function cos 1 ² 1 as →1 tends to zero, therefore it is infinitesimal, therefore you can apply asymptotic formula number two: cos lncos 1 ² 11 P 1 ² 1 Q 2 ² ] P 1 ² 1 Q ^. Now the “small” properties will come in handy for us. We apply them and get: P 1 ² 1 Q 2 1 2 1 ² 1 12 P ² 1 Q 1 2 ² 1 ² 1 1 2 ² 1 . The first thing we did was reveal the numerator - there is the square of the sum. Next, we simply apply the “² small” properties. If you haven't taught them, look at the table I gave earlier. Likewise, P 1 ² 1 Q 1 ² 1 . We apply asymptotic property number 11. We get: ² ] P 1 ² 1 Q ^² 1 ² 1 ² 1 . We finally get cos ln1 1 2 ² 1 as → 1. We can also write our solution like this: lim → cos lnlim → d 1 1 2 ² 1 e . Now you understand why we need these asymptotic formulas! How would you look for this limit differently? Remember, if a function tends to zero, we can always replace it with asymptotic formulas. If it does not tend to zero, but, for example, to some constant or infinity, we do not have the right to use asymptotic formulas!!! Asymptotic formulas are applied only when the function tends to 0! Higher mathematics for dummies. Limit of the function 2011 66 Let's calculate our limit: lim → cos lnlim → d 1 1 2 ² 1 e ¦1 1 1 2 §1. Difficult? No! Confused? Yes! But what can you do, practice is definitely needed here. I think everything will be clear to you in a few minutes. Let's move on to examples. As always, the first one is analyzed in detail, solve the rest of the examples yourself first, and then look at the solution. No. 1. Find the limit: lim → ln1 4 sin3 . Solution: First, let's see if asymptotic formulas can be applied. Let's remember when they can be used? When a function approaches zero. Let's check: lim → ln1 4 ln1 0 lim → sin3 sin0 0 That's right! So we apply the formulas. In this case it is ln1 ¼ ~¼, sin¼~¼. Since the example is very simple, we don’t have to write “small” here. You can use it if you want. Then lim → ln1 4 sin3 lim → 43 43 . As you can see, everything is very simple. No. 2. Find the limit: lim → √ 1 1 . Solution: Since ½ √ 1 1 ¾ →0 and º » →0 as →0, we can apply asymptotic formulas. √ 1 ~1 3 ,. That is, Higher Mathematics for Dummies. Limit of the function 2011 67 lim → √ 1 1 lim → 1 3 1 lim → 3 13 . No. 3. Find the limit: lim !→ 1 cos1 cos sin . Solution: Since º 1 cos1 cos » → 0 and º sin » →0 as →0, we can apply asymptotic formulas. cos ~1 2 ,sin ~. That is, lim !→ 1 cos1 cos sin lim !→ 1 cos!1 1 2 # sin lim !→ 1 cos 2 The example has been simplified, but this is not enough for us. Therefore, since 2 1 cos ! → 0 and º » → 0 as →0, then we can apply asymptotic formulas. cos ~1 2 . lim !→ 1 cos1 cos sin lim !→ 1 cos!1 1 2 # sin lim !→ 1 cos 2 lim !→ 1 p r 1 ! 2 # 2 s u lim !→ 8 v 18 . No. 4. Find the limit: lim → √ 1 2 3 1 . Solution: Since ½√ 1 2 3 1 ¾ → 0 as → 0, we can apply asymptotic formulas. 1 ~1 . Higher mathematics for dummies. Function limit 2011 68 In this case, 1/2. Therefore, this is what we get: lim → √ 1 2 3 1 lim → 1 2 3 2 1 12 lim → 2 3 12 lim → 2 3 7 12 ∙ 2 8 1. No. 5. Find the limit: lim → lnln. Solution: Since º lnln » →0 as →, we can apply asymptotic formulas. ln 1 ¼ ~¼. Thus we get: lim → lnln lim → lnln 1 1 lim → ln1 ln 1 lim → ln 1 lim → ln ln lim → ln lim → ln 1 ] 1^ 2 ln 1 ] 1^ →0at → lim → 1 lim → 1 lim → 1. To be honest, the limit is not the simplest. It’s quite easy to get confused here, so if you, a “dummy,” have taken this limit, then you are far from being the same person you were before reading this book. You are already an average student at a good institute! No. 6. Find the limit: lim → log 1 2 . Solution: Since º log 1 » →0 as → 2, we can apply asymptotic formulas. ln 1 ¼ ~¼. We get: lim → log 1 2 lim → log log 2 2 lim → log 2 2 lim → ln/2 ln2 2 1 ln2 lim → ln/2 2 1 ln2 lim → ln 1 ] 2 1^ 2 1 ln2 lim → 2 1 2 1 2 ∙ ln2 lim → 2 2 1 2 ∙ ln2 . No. 7. Find the limit: Higher mathematics for dummies. Limit of the function 2011 69 lim → sin 1 1 . Solution: Since º sin 1 » →0 as →1, we can apply asymptotic formulas. For sine we have this formula: sin~. Therefore, let's move on to a new variable. Let 1. Then → 0 as →1. The limit becomes equal to ¿lim !→ sin 1 1 Next we use the algebraic identity: 1 4 6 4 1 Thus we find the limit: ¿lim !→ sin 1 1 sin~lim !→ 4 6 4 lim ! → 1 4 6 4 14 . No. 8. Find the limit: lim → lncos √ 1 1 . Solution: Since º lncos » →0 and ½√ 1 1 ¾ →0 as →0, we can apply asymptotic formulas. √ 1 ~1 w ,ln 1 ~. Then the limit can be written as ¿lim → lncos √ 1 1 lim → ln1 cos 1 !1 3 #1 lim → cos 1 /3 3 lim → 1 cos ¦1 cos ~ 2 §3 lim → 2 32 lim → 32 . No. 9. Find the limit: lim → sinsintg! 2 # lncos3 . Higher mathematics for dummies. Function limit 2011 70 Solution: It looks like a terrible example, doesn’t it? Don't worry ☺! We always overcome everything. Let’s also use “² small” in this example so that our answer is definitely correct. Let us write the asymptotic expansion of the numerator using the asymptotic formulas for sine and tangent and the “² small” properties: sinsintg d 2 e = sinsin 2 +² d 2 e ¡ = sin À 2 +² d 2 e +² 2 +² d 2 e ¡ Á = sin 2 +² +² ¡ = sin 2 +² ¡ = 2 +² . Here we used the fact that ² d +² ] ^ e = ²() and ² +² = ²(). Let us now derive the asymptotic expansion of the denominator using the asymptotic formulas for the cosine and logarithm: lncos3 = ln1 − 3 2 +² 3 ¡ = lnn1 +− 9 2 +² ¡o =− 9 2 +² ¡ +² − 9 2 +² ¡ = − 9 2 +² +² = − 9 2 +² . Here we took advantage of the fact that ² 3 = ² ,² − 9 2 +² ¡ = ² ,² +² = ² . Thus, this limit is equal to lim → sinsintg! 2 # lncos 3 = lim → 2 +²() − 9 2 +²() = lim → 12 + ²() − 92 + ²() = 12 +lim → ²() − 92 + lim → ²() = − 19. Here we took advantage of the fact that, by the definition of the symbol “² small”, lim → ² = 0. Higher mathematics for dummies. Function limit 2011 71 From the author: I must say that if you have finally reached this page, then you are far from a dummie! You are already a fully educated person who is well versed in the functions. I have tried to explain this topic to you as clearly as possible. I hope I managed to do this. Next, a large and very important topic awaits you. These are derivatives and differentials. Then, my plans include the topic “indefinite integral”, then “basic theorems on continuous and differentiable functions”. But this is all in the plans for now. I wrote this part and am very pleased with it. Surely, the book contains both grammatical and mathematical errors (loss of sign). Please write to me about this by email... And now you can safely move on to additional chapters ☺. Good luck! Sincerely, Your Viosagmir I.A. [email protected] Higher mathematics for dummies. Function limit 2011 72 Chapter 4. Additional methods. Let's look at additional methods by which we can count our limits. In some cases, these methods are much easier to use than those that we have already gone through. But I must warn you that here you must know how you can and should differentiate the function. Now I will not dwell on this, since this topic is discussed in detail in my second book. So, what makes this L'Hopital method so special? And it is special in that it can reveal uncertainties of the form V 0 0 v W and ∞ ∞ ⁄ . If we remember, we have already gone through many ways to disclose various uncertainties, but there are cases when it is difficult to disclose it, well, or at least inconvenient. But again, L'Hopital's rule does not apply in all cases. The general formulation looks like this: Under certain conditions, the limit of the ratio of functions is equal to the limit of the ratio of their derivatives. Let's look at these conditions ☺. 1. lim → lim → O0or∞ 2. and O are differentiable in a punctured neighborhood 3. O 0 0 in a punctured neighborhood 4. there exists lim → ′ O′ à Then, if the conditions 1 2 3 4 → lim → O lim → ′ O are satisfied ′ . Note that →, and not to some kind of infinity or even zero. What is important to us is that the limit of these functions must be equal to infinity or zero! Many people get confused with this at first, so don’t ignore it ☺. Contents: 1) L'Hopital's rule 2) Taylor series expansion. Part 1 3) Taylor series expansion. Part 2 1. L'Hopital's rule Higher mathematics for dummies. Limit of function 2011 73 I think there is no need to give more theory here. My book is more aimed at practice, so we’ll move on to it now. No. 1. Find the limit lim → +5 3 . Solution: First, let's write out our functions () and O() = +5,O = 3 Now we check our conditions 1. lim → () = lim → +5 = 0,lim → O() = lim → 3 = 0 → ! 2. () and O() are differentiable in a punctured neighborhood. Those. you can take the derivative of these functions at the point = 0 −! 3. O 0 = 3 ≠ 0 in the punctured neighborhood 0 −! 4. there is lim → ′() O′() à = lim → 2 +5 3 v −! Once you get used to it, you won’t waste your precious time checking. I showed you how to do it. Now, I will only check the first point. A parting word for you - check every point! Because anything can happen. lim → +5 3 = 7 00 − ÄÅ 8 = lim → +5 0 3 0 = lim → 2 +5 3 = 7 0 +5 3 − 8 = 53 This is the best solution to this example! 1 – determine the uncertainty; 2 - describe derivatives; 3 – we count derivatives and at the same time see whether () and O() tend to 0; 4 – determine the uncertainty; 5 - write the answer. Easily? Yes! But it takes practice not to get confused. No. 2. Find the limit lim → +4 +7 +3 Solution: = +4 +7 → ∞ as →∞ and O = +3 → ∞ as →∞. Therefore, we can apply L'Hopital's rule ☺. lim → +4 +7 +3 = ∞∞ = lim → +4 +7 0 +3 0 = lim → 3 +8 +7 3 +6 = ∞∞ = lim → 3 +8 +7 ′ 3 +6 ′ = lim → 6 +8 6 +6 = ∞∞ = lim → 6 +8 ′ 6 +6 ′ = lim → 66 = 7 66 8 = 1 Higher mathematics for dummies. Limit of a function 2011 74 Here we had to apply L'Hopital's rule 3 times, because the certainty did not want to go away! Before you start differentiating you should check the conditions on the functions. Here you have checked the conditions 4 times! These are indicated in red - steps where you check the conditions before moving on to the next step. I must say that you probably already realized that this method for this example is clearly not optimal. Here it is better to use what we did for half of this book - take out the numerator and denominator. lim → +4 +7 +3 = lim → ]1 + 4 + 7 ^ ]1 + 3 ^ = lim → 1 + 4 + 7 1 + 3 = 7 1 +0 +0 1 +0 8 = 1 And you can also and do this: lim → +4 +7 +3 = ∞∞ = lim → +4 +7 0 +3 0 = lim → 3 +8 +7 3 +6 = lim → ]3 + 8 + 7 ^ ]3 + 6 ^ = lim → 3 + 8 + 7 3 + 6 = 7 1 +0 +0 3 +0 8 = 33 = 1 That is, at the first step we check for uncertainty and apply L'Hopital's rule, but immediately guess what we need will do this two more times. To save our time, we put the highest degree into the numerator so that we get infinitesimal functions. Why do I spend so much time on this? I want you to understand everything and understand that different methods can be mixed with each other! At the same time, we must not forget about the conditions in each such method. No. 3. Find the limit lim → ln 1 +2lnsin Solution: It is precisely for such cases that we have L'Hopital's rule. How can we solve it differently? Well, maybe some kind of replacement. Since all the conditions are met (check them yourself), we can apply L'Hopital's rule. lim → ln 1 +2lnsin = ∞∞ = lim → ln 0 1 +2lnsin 0 = lim → 1 2 ∙ cos sin = lim → sin 2 cos Didn’t we have a similar example before ☺? In my opinion, this is the first remarkable limit. Let's write it more beautifully: lim → sin 2 cos = 12 ∙ lim → 7! sin #∙ 1 cos 8 = 12 ∙ 1 ∙ lim → 1 cos = 12 Higher mathematics for dummies. Limit of the function 2011 75 Therefore, lim → ln 1 +2lnsin = 12. You see, L'Hopital's rule helps us get to a certain place. And then we apply what we went through with you earlier ☺. Let's move on... No. 4. Find the limit lim → 1 −cos 4 Solution: Since all the conditions are met (check them yourself), we can apply L'Hopital's rule. lim → 1 −cos 4 = 7 00 8 = lim → 1 −cos 4 0 0 = lim → 4sin4 2 = 7 00 8 = lim → (4sin4)′ (2)′ = lim → 16cos 4 2 = 8 Here we applied L'Hopital's rule twice. By the way, this could be solved using the first remarkable limit, after the first application of L'Hopital's rule. We would have it like this: lim → 4sin4 2 = lim → ! sin4 4 #∙ 8 = 8 #5. Find the limit lim → ln Solution: As you can see, we don’t have fractions here. Therefore we cannot apply L'Hopital's rule. But we are savvy, so now we will make the fraction ourselves ☺. ln = ln 1 v Now everything is correct! Check the conditions yourself and make sure that we have the right to apply L'Hopital's rule. lim → ln = 0 ∙ ∞ = lim → ln 1 v = 7 00 8 = lim → ln ′ P 1 v Q ′ = lim → 1 − 1 = −lim → = 0 = 0 No.6. Find the limit lim → ! 1 −1 − 1 ln # Higher mathematics for dummies. Limit of the function 2011 76 Solution: Here, as in the previous example, you need to make a fraction. I hope you know how to add fractions with different denominators ☺. 1 −1 − 1 ln = ln − +1 −1 ln Now everything is correct! Check the conditions yourself and make sure that we have the right to apply L'Hopital's rule. lim → ! 1 −1 − 1 ln #= ∞−∞ = lim → ln − +1 −1 ln = 7 00 8 = lim → (ln − +1)′ (−1 ln)′ = lim → 1 −1 ln + − 1 = lim → 1 − ln + −1 = 7 00 8 = lim → (1 −)′ (ln + −1)′ = lim → −1 1 +ln +1 = 7 − 12 8 = − 12 Here we initially moved on to fractions, then applied L'Hôpital's rule twice in a row. No. 7. Find the limit lim → 1 + Solution: Here you can try to go to the second remarkable limit. We will try to apply Taylor's rule. To do this you need to make a fraction. Let's do it quite cleverly - let's denote 1 + for. That is, 1 + = →ln = 1 ∙ ln 1 + = ln 1 + Now we use a very useful property at the moment: Since Ä is a continuous function, then lnlim → = lim → ln I bet half of you didn’t understand anything ☺ . In short, in this example we move from one function to another, without forgetting to change the limits. º → 0at | →∞priln » Right? Yes! Remember the graph of the logarithm. Accordingly, having changed the limits, we begin to look for the limit using L'Hopital's rule. lim → ln = lim → ln 1 + = ∞∞ = lim → ln 1 + ′ ′ = lim → 2 1 + = ∞∞ = lim → (2)′ 1 + ′ = lim → 2 2 = 0 Now don’t forget to go to the reverse repartitions! Those. We get Higher Mathematics for Dummies. Limit of the function 2011 77 lim → = orlim → 1 + = 1 Interesting example ☺? The most important thing is that you understand that the same example can be solved in different ways, and not just one. No. 8. Find the limit lim → −2arctg ln Solution: We cannot apply L'Hopital's rule since there is no fraction. Therefore, we do it −2arctg ln = −2arctg 1 ln You check 4 properties and understand that L'Hopital's rule can be applied. lim → −2arctg ln = lim → −2arctg 1 ln = () 7 00 8 = lim → −2arctg ′ ] 1 ln ^ ′ = lim → − 2 1 + − 1 ln = lim → 2 ln 1 + = () ∞ ∞ = lim → (2 ln)′ (1 +)′ = lim → 2ln +4ln 2 = () ∞∞ = lim → (2ln +4ln)′ (2)′ = lim → 4 ∙ ln + 4 2 = 2 lim → ln +1 = () ∞∞ = lim → (ln +1)′ ′ = lim → 1 v 1 = lim → 1 = 0 We used as many as four L'Hopital rules here! It looks like a beautiful solution, of course ☺. I want to tell you that such examples are not solved in every university. I want you to decide these things! And they were not, so to speak, “dummies.” No. 9. Find the limit lim → arcsin 1 Solution: This is also a little tricky ☺. We need to use the property of the logarithm arcsin 1 = 1∙23/.& = 23/.& /1 How did we do this? It's simple. There is such a formula: = We just use it and get Higher mathematics for dummies. Function limit 2011 78 arcsin 1 = 4 23/.& 5 = 1∙23/.& = 23/.& /1 That is, we can write everything like this: lim → arcsin 1 = &" → 6 1∙23/ .& 7 = &" → 9 23/.& /1: = ;< = &" → 9 23/.& 0 (/1)0: = &" → 9 .& √ ∙23/.& : = &" → 9 ∙ : = = 1 Вот этот пример уже не шутка. Это полноценный, выше среднего уровня, пример! №10. Найти предел lim → ctg Решение: Здесь делаем то же самое, что и в предыдущем примере. Таким образом у нас получается lim → ctg = &" → /1 = ; < = &" → /1 0 ()0 = &" → .& ∙/1 = &" → .&∙/. = = 1 №11. Найти предел lim → −sin +sin Решение: Здесь нельзя применять правило Лопиталя! Проверьте все условия и поймите, что я говорю все верно ☺! Здесь нужно вычислять предел вот так lim → −sin +sin = lim → ]1 − sin ^ ]1 + sin ^ = lim → 1 − sin 1 + sin = 1 №12. Найти предел lim → ! sin # Решение: lim → ! sin # = &" → = .& >∙ = &" → .& = ; < = &" → .& ()0 = &" → /. = ; < = &" → = Высшая математика для чайников. Предел функции 2011 год 79 На этом мы заканчиваем правило Лопиталя. Запомните одну важную вещь! Не стоит применять это правило езде и вся. Сначала определите, а нужно ли вообще его здесь применять? Когда у вас логарифмы, синусы, корни, то оно может помочь. Но если у вас простые выражения, оно может только затруднить вашу работу. Так что никуда не спешите ☺. Высшая математика для чайников. Предел функции 2011 год 80 В данном разделе мы рассмотрим предел функции вида O ⁄ . Что такое разложение ряда Тейлора и все его подробности я рассказывать не буду, так как это все написано в моей второй книге. В данном разделе я на примерах объясню принцип данной работы. В таблице представлены основные разложения по формуле Тейлора при условии, что →0. Их можно не запоминать, просто распечатайте и пользуйтесь ими. А сейчас мы разберем метод Тейлора на конкретных примерах. Я называю данные примеры crash-примерами. Сейчас поймете, почему именно такое название ☺. №1. Найти предел lim → cos arctg ln 1 Решение: Так как в знаменателе одна функция, то представим ее формулой Маклорена до остаточного члена ² , то есть sin 6 & 120 & 6 cos 1 2 & 24 & 6 7 Y & 6 & 120 & 6 " Y 1 & 2 & 24 & 6 tg & 3 & 2 15 & 6 8 Y 3 & 2 15 & 6 arcsin & 6 & 3 40 & 6 arctg 3 & 5 & 6 ln 1 & 2 & 3 4 & 6 ln 1 2 3 4 & 6 ln > & Z 1 & E 6 & 3 40 & 6 1 1 & 1 & & & 6 1 1 1 & & & & & 6 √ 1 & 1 & 2 8 & 16 & 6 2. Taylor series expansion. Part 1 Higher mathematics for dummies. Limit of the function 2011 81 O = − ∙ +² = − +²() The denominator of the fraction can be easily represented as a Maclaurin series. We don’t need all the members, so we take the very first, non-zero one. Now let's look at the numerator. Since we expanded the denominator to the remainder term ², then we must expand the numerator to exactly the same remainder term. cos = 1 − 2 +² → ∙ cos = − 2 +² arctg = − 3 +² As you can see, we expand cos to the remainder term ² , since we already know that we multiply cos by and it will give us the remainder term ² . As a result, here is our decomposed numerator: = − 2 +² − − 3 +² ¡ = − 6 +² Then lim → () O() = lim → − 6 +² − +²() = 16 Here we are and calculated the first limit ☺. Confused? Yes. But with the help of Taylor series, very complex and “impenetrable” limits can be calculated. Once you know how to do this, you will spend a lot of time looking for the limit, but you will eventually find it! You will win ☺. No. 2. Find the limit lim → sin] 1 − ^ +ln(1 −) − 2 tan(Åℎ) −arctg Solution: First, consider the denominator and try to find the O() function. To do this, let us expand our functions tg(Åℎ) and arctg. Now the question arises, to what remainder term should we expand? Well, first, let's try before ²(). Åℎ = +²() O = +², where = Åℎ Now let’s substitute and find O(Åℎ) O Åℎ = +² +² P +² Q = +²() Higher mathematics for dummies. Limit of the function 2011 82 But let's look at the numerator. There, the remainder term in the expansion will be greater than ²(). As I already said, the remainder term must be the same everywhere. Therefore we will have to expand to ². Åℎ = + 3! +² O = + 3 +² ,where = Åℎ Now let’s substitute and find O(Åℎ) O Åℎ = + 3 +² = + 3! +²¡ + d + 3! +² e 3 +² + 3! +² ¡ Now let’s pay attention to the second term, i.e. on d+3! +² e 3 If we open the brackets in the numerator, we get + 2 + % 4 + 19 +² But! We don't need it, we need it, as we agreed before. Therefore we can get rid of the terms 2 + % 4 + 19 Because they give us ². I repeat once again, if we decided that in our example the remainder term will be presented in the form ², then it must be exactly like this in each term and not otherwise! Accordingly, we can write this: O Åℎ = + 3 +² = + 3! +² ¡ + P +² Q 3 +² = + 2 +² Let's expand the second term in the denominator. We already have it in the table arctg = − 3 +² Thus, the denominator function O() is expanded as follows: O = + 2 +² ¡ − − 3 +² ¡ = 5 6 +² Higher mathematics for dummies. Limit of the function 2011 83 Now let's move on to the numerator. First, let's look at 1 - We have a formula for the type of fraction 1 1 - We'll do it cleverly. Let's expand the fraction to the remainder term ², since when we then multiply by we get the estimate ². And that’s exactly what we need! 1 1 − = 1 + + +² Then, when multiplied by we get 1 − = P 1 + + +² Q = + + +² Let us expand sin, where = 1 − v . We also know this formula (in the table). sin = − 3! +² Here we also expanded to ², since we don’t have any multiplications by sin. Now let's substitute everything under and get sin] 1 − ^ = P + + +² Q − P + + +² Q 3! +² ] P + + +² Q ^ Now consider our fraction P + + +² Q 3! Pay attention to the numerator. If we open the brackets, our estimate will increase significantly, and we don't want that. We need the rating to remain ². What to do? Get rid of the rest of the members! Thus, the fraction will take a slightly different form: P +² Q 3! Of course, if you want, you can open all the brackets P + + +² Q , E and then throw out all the ones whose degree is greater than 3. But you will be tired of doing this, so throw them out right away! So, this is what we get: Higher mathematics for dummies. Limit of the function 2011 84 sin] 1 − ^ = P + + +² Q − P +² Q 3! +² = + + +² − +² 3! = + + 5 6 +² Let's consider the second term in the numerator, that is, ln(1 −) Thank God, we already have its expansion in the table ln(1 −) = − − 2 − 3 +² Total, we can write down our () function = + + 5 6 +² ¡ +− − 2 − 3 +² ¡ − 2 = 2 +² Now we have the functions () and O() expanded. We can find our limit lim → () O() = lim → 2 +² 5 6 +² = 35 We have found the limit! I want to say that this is the highest level! This is not a “teapot” and not an “average” one. This is a mega-student who can do a lot. Gentlemen, increase your self-esteem and feel superior to others by solving such examples ☺. Personally, I sincerely hope that you will understand (or maybe already understand) everything that I am telling you. Well!? Let's continue to conquer the heights of mathematics ☺! No. 3. lim → O P Q − ln Eℎ arctg(cos) −tg Solution: Beauty, isn't it ☺? It’s okay, we’ve completed the previous one, let’s conquer this one too! We will present it to accuracy ², as in previous issues. Let's try to derive the O() function. To do this, consider cos (we know its expansion) cos = 1 − 2 +² The remainder term is presented in the form ², since we multiply by cos, which gives us our best estimate ². Higher mathematics for dummies. Function limit 2011 85 cos = 1 − 2 +² ¡ = − 2 +² Now let’s expand arctg , where = cos (also according to the table) EO = − 3 +² Then we can expand arctg(cos) arctg(cos) = − 2 +² ¡ − d − 2 +² e 3 +² n − 2 +² ¡ o If we pay attention to the numerator of the second fraction, that is, − 2 +² ¡ , then we will immediately notice that when opening the parentheses, we will not get ² in any way. The degree y will be much higher. Therefore, we get rid of the terms we don’t need and get arctg(cos) = − 2 +² ¡ − +² 3 +² = − 5 6 +² We just have to expand the last term in the denominator O = + 3 +² Thus, we have collected all the necessary we need data in order to find the O() function. O = arctan(cos) −tg = − 5 6 +² ¡ − + 3 +² ¡ = − 7 6 +²() Great! We were able to represent the denominator to within ². Therefore, we can safely move on to the numerator. We need to expand O P Q − ln Eℎ As you probably already realized, we start with internal functions. So, let’s break it down first! , where = − . ! = 1 + +²() Higher mathematics for dummies. Limit of the function 2011 86 As you can see, we expand with an accuracy of ²(), since it will give us an accuracy of ², and − ². = 1 − +² = P 1 − +² Q = − +² Now let’s expand O, where = . O = + 3 +² Substitute and get O P Q = P − +² Q + P − +² Q 3 +² ] P − +² Q ^ Consider the numerator of the second fraction P − +² Q If we open the brackets, then we already have there will be no precision ², so we simply get rid of the other members. O P Q = P − +² Q + P +² Q 3 +² = − 2 3 +² Great! We were able to imagine one term. Now let's look at the second ln Eℎ There is also a trick here. Since we are dividing by, we need to present the numerator with an accuracy of ², so that when dividing, the accuracy of the entire fraction would be ². ln(Eℎ) = 2ln(Eℎ) Here we simply applied the property of the logarithm. Eℎ = 1 + 2 + 24 +² Now let’s expand ln(+1), where = Eℎ −1. We expand ln(+1) since we do not have expansion formulas for ln. = Eℎ −1 − with this we compensate for our unity. Higher mathematics for dummies. Function limit 2011 87 ln(+1) = − 2 + 3 − 4 +² = 1 + 2 + 24 +² ¡ − d 1 + 2 + 24 +² e 2 + d 1 + 2 + 24 +² e 3 − d 1 + 2 + 24 +² e 4 +² n1 + 2 + 24 +² ¡ o Well, then. Here we must discard all terms so that the estimate does not increase, but also remains at the level ². This is what we end up with ln(Eℎ) = 2ln(Eℎ) = 2ln1 + 2 + 24 +² ¡ = 2 p q q r 2 + 24 +² − d 2 +² e 2 +² s t t u = 2 2 + 24 + ² − 8 +² ¡ = − 6 +² Thus, we can write our function () = − 2 3 +² ¡ − 1 − 6 +² ¡ = − 2 +² From here we can find the limit lim → () O( ) = lim → − 2 +² − 7 6 +²() = 37 Higher mathematics for dummies. Limit of a function 2011 88 In this topic we will look at the limit of a function of the form? . Just like in the last section, let's look at everything using examples. No. 1. Find the limit of the function lim → d √1 cos e Solution: Let us write down the expansion of the function. This is easy to do, since we have all the expansions in the table. √1 cos 1 12 18 ² 1 12 1 24 ² d 1 12 18 ² eÆ 1 d 12 1 24 ² e 2 ² ¡ ² Ç 1 2 8 ² ¡1 2 5 24 ² ¡1 6 ² From here it is easy to find the limit lim → ? lim → 1 6 ² ¡ / We have already covered how to calculate the second wonderful limit, so I won’t waste time on it now. 3. Taylor series expansion. Part 2

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When calculating limits, one should take into account the following basic rules:

1. The limit of the sum (difference) of functions is equal to the sum (difference) of the limits of the terms:

2. The limit of a product of functions is equal to the product of the limits of the factors:

3. The limit of the ratio of two functions is equal to the ratio of the limits of these functions:

.

4. The constant factor can be taken beyond the limit sign:

.

5. The limit of a constant is equal to the constant itself:

6. For continuous functions, the limit and function symbols can be swapped:

.

Finding the limit of a function should begin by substituting the value into the expression for the function. Moreover, if the numerical value 0 or ¥ is obtained, then the desired limit has been found.

Example 2.1. Calculate the limit.

Solution.

.

Expressions of the form , , , , , are called uncertainties.

If you get an uncertainty of the form , then to find the limit you need to transform the function so as to reveal this uncertainty.

Uncertainty of form is usually obtained when the limit of the ratio of two polynomials is given. In this case, to calculate the limit, it is recommended to factor the polynomials and reduce them by a common factor. This multiplier is zero at the limit value X .

Example 2.2. Calculate the limit.

Solution.

Substituting , we get uncertainty:

.

Let's factor the numerator and denominator:

;

Let's reduce by a common factor and get

An uncertainty of the form is obtained when the limit of the ratio of two polynomials is given at . In this case, to calculate it, it is recommended to divide both polynomials by X in the senior degree.

Example 2.3. Calculate the limit.

Solution. When substituting ∞, we obtain an uncertainty of the form , so we divide all terms of the expression by x 3.

.

It is taken into account here that .

When calculating the limits of a function containing roots, it is recommended to multiply and divide the function by its conjugate.

Example 2.4. Calculate limit

Solution.

When calculating limits to reveal uncertainty of the form or (1) ∞, the first and second remarkable limits are often used:

Many problems associated with the continuous growth of some quantity lead to the second remarkable limit.

Let's consider the example of Ya. I. Perelman, giving an interpretation of the number e in the compound interest problem. In savings banks, interest money is added to the fixed capital annually. If the accession is made more often, then the capital grows faster, since a larger amount is involved in the formation of interest. Let's take a purely theoretical, very simplified example.

Let 100 deniers be deposited in the bank. units based on 100% per annum. If interest money is added to the fixed capital only after a year, then by this period 100 den. units will turn into 200 monetary units.

Now let's see what 100 denize will turn into. units, if interest money is added to fixed capital every six months. After six months, 100 den. units will grow by 100 × 1.5 = 150, and after another six months - by 150 × 1.5 = 225 (den. units). If the accession is done every 1/3 of the year, then after a year 100 den. units will turn into 100 × (1 +1/3) 3 "237 (den. units).

We will increase the terms for adding interest money to 0.1 year, to 0.01 year, to 0.001 year, etc. Then out of 100 den. units after a year it will be:

100 × (1 +1/10) 10 » 259 (den. units),

100 × (1+1/100) 100 » 270 (den. units),

100 × (1+1/1000) 1000 » 271 (den. units).

With an unlimited reduction in the terms for adding interest, the accumulated capital does not grow indefinitely, but approaches a certain limit equal to approximately 271. The capital deposited at 100% per annum cannot increase by more than 2.71 times, even if the accrued interest were added to the capital every just a second because

Example 2.5. Calculate the limit of a function

Solution.

Example 2.6. Calculate the limit of a function .

Solution. Substituting we get the uncertainty:

.

Using the trigonometric formula, we transform the numerator into a product:

As a result we get

Here the second remarkable limit is taken into account.

Example 2.7. Calculate the limit of a function

Solution.

.

To reveal uncertainty of the form or, you can use L'Hopital's rule, which is based on the following theorem.

Theorem. The limit of the ratio of two infinitesimal or infinitely large functions is equal to the limit of the ratio of their derivatives

Note that this rule can be applied several times in a row.

Example 2.8. Find

Solution. When substituting , we have an uncertainty of the form . Applying L'Hopital's rule, we get

Continuity of function

An important property of a function is continuity.

Definition. The function is considered continuous, if a small change in the value of the argument entails a small change in the value of the function.

Mathematically this is written as follows: when

By and is meant the increment of variables, that is, the difference between the subsequent and previous values: , (Figure 2.3)

Figure 2.3 – Increment of variables

From the definition of a function continuous at the point it follows that . This equality means that three conditions are met:

Solution. For function the point is suspicious for a discontinuity, let's check this and find one-sided limits

Hence, , Means - break point

Derivative of a function

We figured out the basic elementary functions.

When moving to functions of a more complex type, we will certainly encounter the appearance of expressions whose meaning is not defined. Such expressions are called uncertainties.

Let's list everything main types of uncertainties: zero divided by zero (0 by 0), infinity divided by infinity, zero multiplied by infinity, infinity minus infinity, one to the power of infinity, zero to the power of zero, infinity to the power of zero.

ALL OTHER EXPRESSIONS OF UNCERTAINTY ARE NOT AND TAKE A COMPLETELY SPECIFIC FINITE OR INFINITE VALUE.

Uncover uncertainty allows:

• simplification of the type of function (transformation of expressions using abbreviated multiplication formulas, trigonometric formulas, multiplication by conjugate expressions followed by reduction, etc.);
• use of remarkable limits;
• application of L'Hopital's rule;
• using the replacement of an infinitesimal expression with its equivalent (using a table of equivalent infinitesimals).

Let's group the uncertainties into uncertainty table. For each type of uncertainty we associate a method for its disclosure (method of finding the limit).

This table, together with the table of limits of basic elementary functions, will be your main tools in finding any limits.

Let's give a couple of examples when everything works out immediately after substituting the value and uncertainty does not arise.

Example.

Calculate limit

Solution.

Substitute the value:

And we immediately received an answer.

Answer:

Example.

Calculate limit

Solution.

We substitute the value x=0 into the base of our exponential power function:

That is, the limit can be rewritten as

Now let's take a look at the indicator. This is a power function. Let us turn to the table of limits for power functions with a negative exponent. From there we have And , therefore, we can write .

Based on this, our limit will be written as:

We turn again to the table of limits, but for exponential functions with a base greater than one, from which we have:

Answer:

Let's look at examples with detailed solutions Uncovering uncertainties by transforming expressions.

Very often the expression under the limit sign needs to be slightly transformed to get rid of uncertainties.

Example.

Calculate limit

Solution.

Substitute the value:

We have arrived at uncertainty. We look at the uncertainty table to select a solution method. Let's try to simplify the expression.

Answer:

Example.

Calculate limit

Solution.

Substitute the value:

We came to uncertainty (0 to 0). We look at the uncertainty table to choose a solution method and try to simplify the expression. Let's multiply both the numerator and the denominator by the expression conjugate to the denominator.

For the denominator the conjugate expression will be

We multiplied the denominator so that we could apply the abbreviated multiplication formula - difference of squares and then reduce the resulting expression.

After a series of transformations, the uncertainty disappeared.

Answer:

COMMENT: For limits of this type, the method of multiplying by conjugate expressions is typical, so feel free to use it.

Example.

Calculate limit

Solution.

Substitute the value:

We have arrived at uncertainty. We look at the uncertainty table to choose a solution method and try to simplify the expression. Since both the numerator and the denominator vanish at x = 1, then if these expressions can be reduced (x-1) and the uncertainty will disappear.

Let's factorize the numerator:

Let's factorize the denominator:

Our limit will take the form:

After the transformation, the uncertainty was revealed.

Answer:

Let's consider limits at infinity from power expressions. If the exponents of the power expression are positive, then the limit at infinity is infinite. Moreover, the greatest degree is of primary importance; the rest can be discarded.

Example.

Example.

If the expression under the limit sign is a fraction, and both the numerator and the denominator are power expressions (m is the power of the numerator, and n is the power of the denominator), then when an uncertainty of the form infinity to infinity arises, in this case uncertainty is revealed dividing both the numerator and denominator by

Example.

Calculate limit