For those who want to learn how to find the limits in this article, we will tell you about it. We will not delve into the theory, usually teachers give it at lectures. So the "boring theory" should be outlined in your notebooks. If this is not the case, then you can read textbooks taken from the library of the educational institution or on other Internet resources.

So, the concept of a limit is quite important in studying a course in higher mathematics, especially when you come across integral calculus and understand the relationship between the limit and the integral. The current article will look at simple examples, as well as ways to solve them.

Examples of solutions

Example 1

Calculate a) $ \ lim_ (x \ to 0) \ frac (1) (x) $; b) $ \ lim_ (x \ to \ infty) \ frac (1) (x) $

Solution

a) $$ \ lim \ limits_ (x \ to 0) \ frac (1) (x) = \ infty $$ b) $$ \ lim_ (x \ to \ infty) \ frac (1) (x) = 0 $$ We are often sent these limits with a request to help solve. We decided to highlight them as a separate example and explain that these limits must be simply remembered, as a rule. If you can't solve your problem, then send it to us. We will provide a detailed solution. You will be able to familiarize yourself with the course of the calculation and get information. This will help you get credit from your teacher in a timely manner!

Answer

$$ \ text (a)) \ lim \ limits_ (x \ to 0) \ frac (1) (x) = \ infty \ text (b)) \ ​​lim \ limits_ (x \ to \ infty) \ frac (1 ) (x) = 0 $$

What to do with uncertainty like: $ \ bigg [\ frac (0) (0) \ bigg] $

Example 3

Solve $ \ lim \ limits_ (x \ to -1) \ frac (x ^ 2-1) (x + 1) $

Solution

As always, we start by substituting the value of $ x $ in the expression under the limit sign. $$ \ lim \ limits_ (x \ to -1) \ frac (x ^ 2-1) (x + 1) = \ frac ((- 1) ^ 2-1) (- 1 + 1) = \ frac ( 0) (0) $$ What's next? What should be the result? Since this is uncertainty, this is not an answer yet and we continue the calculation. Since we have a polynomial in the numerators, we factor it into factors, using the formula familiar to everyone since school $$ a ^ 2-b ^ 2 = (a-b) (a + b) $$. Do you remember? Fine! Now go ahead and apply it with the song :) We get that the numerator $ x ^ 2-1 = (x-1) (x + 1) $ We continue to solve given the above transformation: $$ \ lim \ limits_ (x \ to -1) \ frac (x ^ 2-1) (x + 1) = \ lim \ limits_ (x \ to -1) \ frac ((x-1) (x + 1)) (x + 1) = $$ $$ = \ lim \ limits_ (x \ to -1) (x-1) = - 1-1 = -2 $$

Answer

$$ \ lim \ limits_ (x \ to -1) \ frac (x ^ 2-1) (x + 1) = -2 $$

Let us push the limit in the last two examples to infinity and consider the uncertainty: $ \ bigg [\ frac (\ infty) (\ infty) \ bigg] $

Example 5

Evaluate $ \ lim \ limits_ (x \ to \ infty) \ frac (x ^ 2-1) (x + 1) $

Solution

$ \ lim \ limits_ (x \ to \ infty) \ frac (x ^ 2-1) (x + 1) = \ frac (\ infty) (\ infty) $ What to do? How to be? Don't panic, because the impossible is possible. It is necessary to put the x outside the brackets in both the numerator and the denominator, and then reduce it. Then try to calculate the limit. Trying ... $$ \ lim \ limits_ (x \ to \ infty) \ frac (x ^ 2-1) (x + 1) = \ lim \ limits_ (x \ to \ infty) \ frac (x ^ 2 (1- \ frac (1) (x ^ 2))) (x (1+ \ frac (1) (x))) = $$ $$ = \ lim \ limits_ (x \ to \ infty) \ frac (x (1- \ frac (1) (x ^ 2))) ((1+ \ frac (1) (x))) = $$ Using the definition from Example 2 and substituting infinity for x, we get: $$ = \ frac (\ infty (1- \ frac (1) (\ infty))) ((1+ \ frac (1) (\ infty))) = \ frac (\ infty \ cdot 1) (1+ 0) = \ frac (\ infty) (1) = \ infty $$

Answer

$$ \ lim \ limits_ (x \ to \ infty) \ frac (x ^ 2-1) (x + 1) = \ infty $$

Algorithm for calculating limits

So, let's briefly summarize the analyzed examples and compose an algorithm for solving the limits:

1. Substitute point x in the expression following the limit sign. If you get a certain number, or infinity, then the limit is completely solved. Otherwise, we have ambiguity: "divide zero by zero" or "divide infinity by infinity" and proceed to the next paragraphs of the instruction.
2. To eliminate the ambiguity "zero divided by zero" you need to factor the numerator and denominator into factors. Reduce similar ones. Substitute point x in the expression under the limit sign.
3. If the uncertainty is "infinity divided by infinity", then we take out both in the numerator and in the denominator of x to the greatest degree. Reducing the x's. Substitute the x values ​​from under the limit into the remaining expression.

In this article, you learned the basics of solving limits commonly used in the Math Analysis course. Of course, these are not all types of problems offered by examiners, but only the simplest limits. In the following articles we will talk about other types of tasks, but first you need to learn this lesson in order to move on. We will discuss what to do if there are roots, degrees, we will study infinitesimal equivalent functions, wonderful limits, L'Hôpital's rule.

If you can't figure out the limits on your own, then don't panic. We are always happy to help!

When calculating limits, consider 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 the 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 out of the limit sign:

.

5. The limit of the constant is equal to the most constant:

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

.

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

Example 2.1. Calculate the limit.

Solution.

.

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

If an uncertainty of the form is obtained, then in order to find the limit, it is necessary to transform the function so as to reveal this uncertainty.

Uncertainty of kind is usually obtained when the limit of the ratio of two polynomials is given. In this case, it is recommended to factor the polynomials and cancel them by a common factor to calculate the limit. This factor is equal to zero at the limit value NS .

Example 2.2. Calculate the limit.

Solution.

Substituting, we get the uncertainty:

.

Let's factorize the numerator and denominator:

;

Reduce by a common factor and get

.

The uncertainty of the form is obtained when the limit of the ratio of two polynomials is given at. In this case, for the calculation, it is recommended to divide both polynomials by NS in the senior degree.

Example 2.3. Calculate the limit.

Solution. Substitution ∞ gives an uncertainty of the form, so we divide all terms of the expression by x 3.

.

It is taken into account that.

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

Example 2.4. Calculate Limit

Solution.

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

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

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

Let the bank put 100 den. units at the rate of 100% per annum. If interest money will be added to the fixed capital only after a year, then by this date 100 den. units will turn into 200 monetary units.

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

We will speed up the terms for joining interest-bearing money to 0.1 years, to 0.01 years, to 0.001 years, etc. Then out of 100 den. units after a year it will turn out:

100 × (1 +1/10) 10 "259 (monetary units),

100 × (1 + 1/100) 100 * 270 (monetary units),

100 × (1 + 1/1000) 1000 * 271 (monetary units).

With an unlimited reduction in the terms of interest attachment, the accrued capital does not grow infinitely, but approaches a certain limit, equal to approximately 271. The capital allocated at 100% per annum cannot increase by more than 2.71 times, even if the accrued interest was added to the capital each 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, convert the numerator to product:

As a result, we get

Here, a second remarkable limit is taken into account.

Example 2.7. Calculate the limit of a function

Solution.

.

To disclose the uncertainty of the form or, you can use the L'Hôpital 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'Hôpital's rule, we get

Continuity of function

Continuity is an important property of a function.

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, it is written as follows: for

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

Figure 2.3 - Increment of variables

It follows from the definition of a function continuous at a point that ... This equality means the fulfillment of three conditions:

Solution. For function the point is suspicious of a break, check it, find one-sided limits

Hence, , means - removable discontinuity point

Derivative of a function

Limits give all math students a lot of hassle. To solve the limit, sometimes you have to use a lot of tricks and choose from a variety of solution methods exactly the one that is suitable for a specific example.

In this article we will not help you understand the limits of your capabilities or comprehend the limits of control, but we will try to answer the question: how to understand the limits in higher mathematics? Understanding comes with experience, so at the same time we will give several detailed examples of solving the limits with explanations.

Limit concept in mathematics

The first question: what is this limit and what is the limit? We can talk about the limits of numerical sequences and functions. We are interested in the concept of the limit of a function, since it is with them that students most often encounter. But first, the most general definition of a limit:

Let's say there is some variable. If this value in the process of change is unlimitedly approaching a certain number a , then a Is the limit of this value.

For a function defined in a certain interval f (x) = y the limit is such a number A , to which the function tends at NS tending to a certain point a ... Point a belongs to the interval on which the function is defined.

It sounds cumbersome, but it's very simple to write:

Lim- from English limit is the limit.

There is also a geometric explanation for the definition of the limit, but here we will not go into theory, since we are more interested in the practical than the theoretical side of the issue. When we say that NS tends to some value, this means that the variable does not take the value of the number, but is infinitely close to it.

Let's give a concrete example. The challenge is to find the limit.

To solve this example, substitute the value x = 3 into a function. We get:

By the way, if you are interested in basic operations on matrices, read a separate article on this topic.

In the examples NS can strive for any value. It can be any number or infinity. Here is an example when NS tends to infinity:

It is intuitively clear that the larger the number in the denominator, the lower the value the function will take. So, with unlimited growth NS meaning 1 / x will decrease and approach zero.

As you can see, in order to solve the limit, you just need to substitute the value to strive for into the function NS ... However, this is the simplest case. Finding the limit is often not so obvious. Uncertainties such as 0/0 or infinity / infinity ... What to do in such cases? To resort to tricks!

Uncertainties within

Uncertainty of the form infinity / infinity

Let there be a limit:

If we try to substitute infinity into the function, we get infinity in both the numerator and denominator. In general, it is worth saying that there is a certain element of art in resolving such uncertainties: it must be noted how a function can be transformed in such a way that the uncertainty disappears. In our case, we divide the numerator and denominator by NS in the senior degree. What happens?

From the example already considered above, we know that the terms containing x in the denominator will tend to zero. Then the solution to the limit is:

To disclose ambiguities like infinity / infinity divide the numerator and denominator by NS to the highest degree.

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Another type of uncertainty: 0/0

As always, substitution in the value function x = -1 gives 0 in the numerator and denominator. Look a little more closely and you will notice that we have a quadratic equation in the numerator. Find the roots and write:

Let's shorten and get:

So, if you are faced with an uncertainty like 0/0 - factor out the numerator and denominator.

To make it easier for you to solve examples, we give a table with the limits of some functions:

L'Hôpital's rule within

Another powerful technique for eliminating both types of uncertainty. What is the essence of the method?

If there is uncertainty in the limit, we take the derivative of the numerator and denominator until the uncertainty disappears.

L'Hôpital's rule looks like this:

An important point : the limit in which instead of the numerator and denominator are derivatives of the numerator and denominator, must exist.

And now for a real example:

Typical uncertainty 0/0 ... Let's take the derivatives of the numerator and denominator:

Voila, ambiguity is resolved quickly and elegantly.

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The first remarkable limit is called the following equality:

\ begin (equation) \ lim _ (\ alpha \ to (0)) \ frac (\ sin \ alpha) (\ alpha) = 1 \ end (equation)

Since for $ \ alpha \ to (0) $ we have $ \ sin \ alpha \ to (0) $, it is said that the first remarkable limit reveals an uncertainty of the form $ \ frac (0) (0) $. Generally speaking, in formula (1), instead of the variable $ \ alpha $ under the sine sign and in the denominator, any expression can be located, as long as two conditions are met:

1. Expressions under the sine sign and in the denominator simultaneously tend to zero, i.e. there is an uncertainty of the form $ \ frac (0) (0) $.
2. The expressions under the sine sign and in the denominator are the same.

Consequences from the first remarkable limit are also often used:

\ begin (equation) \ lim _ (\ alpha \ to (0)) \ frac (\ tg \ alpha) (\ alpha) = 1 \ end (equation) \ begin (equation) \ lim _ (\ alpha \ to (0) ) \ frac (\ arcsin \ alpha) (\ alpha) = 1 \ end (equation) \ begin (equation) \ lim _ (\ alpha \ to (0)) \ frac (\ arctg \ alpha) (\ alpha) = 1 \ end (equation)

Eleven examples have been solved on this page. Example No. 1 is devoted to the proof of formulas (2) - (4). Examples # 2, # 3, # 4 and # 5 contain solutions with detailed comments. Examples # 6-10 contain solutions with almost no comments, since detailed explanations were given in the previous examples. The solution uses some of the trigonometric formulas that can be found.

Note that the presence of trigonometric functions, coupled with the uncertainty $ \ frac (0) (0) $, does not mean that the first remarkable limit must be applied. Sometimes simple trigonometric transformations are enough - for example, see.

Example # 1

Prove that $ \ lim _ (\ alpha \ to (0)) \ frac (\ tg \ alpha) (\ alpha) = 1 $, $ \ lim _ (\ alpha \ to (0)) \ frac (\ arcsin \ alpha ) (\ alpha) = 1 $, $ \ lim _ (\ alpha \ to (0)) \ frac (\ arctg \ alpha) (\ alpha) = 1 $.

a) Since $ \ tg \ alpha = \ frac (\ sin \ alpha) (\ cos \ alpha) $, then:

$$ \ lim _ (\ alpha \ to (0)) \ frac (\ tg (\ alpha)) (\ alpha) = \ left | \ frac (0) (0) \ right | = \ lim _ (\ alpha \ to (0)) \ frac (\ sin (\ alpha)) (\ alpha \ cos (\ alpha)) $$

Since $ \ lim _ (\ alpha \ to (0)) \ cos (0) = 1 $ and $ \ lim _ (\ alpha \ to (0)) \ frac (\ sin \ alpha) (\ alpha) = 1 $ , then:

$$ \ lim _ (\ alpha \ to (0)) \ frac (\ sin (\ alpha)) (\ alpha \ cos (\ alpha)) = \ frac (\ displaystyle \ lim _ (\ alpha \ to (0)) \ frac (\ sin (\ alpha)) (\ alpha)) (\ displaystyle \ lim _ (\ alpha \ to (0)) \ cos (\ alpha)) = \ frac (1) (1) = 1. $$

b) Let's make the substitution $ \ alpha = \ sin (y) $. Since $ \ sin (0) = 0 $, then from the condition $ \ alpha \ to (0) $ we have $ y \ to (0) $. In addition, there is a neighborhood of zero in which $ \ arcsin \ alpha = \ arcsin (\ sin (y)) = y $, so:

$$ \ lim _ (\ alpha \ to (0)) \ frac (\ arcsin \ alpha) (\ alpha) = \ left | \ frac (0) (0) \ right | = \ lim_ (y \ to (0)) \ frac (y) (\ sin (y)) = \ lim_ (y \ to (0)) \ frac (1) (\ frac (\ sin (y)) ( y)) = \ frac (1) (\ displaystyle \ lim_ (y \ to (0)) \ frac (\ sin (y)) (y)) = \ frac (1) (1) = 1. $$

The equality $ \ lim _ (\ alpha \ to (0)) \ frac (\ arcsin \ alpha) (\ alpha) = 1 $ is proved.

c) Let's make the substitution $ \ alpha = \ tg (y) $. Since $ \ tg (0) = 0 $, the conditions $ \ alpha \ to (0) $ and $ y \ to (0) $ are equivalent. In addition, there is a neighborhood of zero in which $ \ arctg \ alpha = \ arctg \ tg (y)) = y $, therefore, based on the results of item a), we will have:

$$ \ lim _ (\ alpha \ to (0)) \ frac (\ arctg \ alpha) (\ alpha) = \ left | \ frac (0) (0) \ right | = \ lim_ (y \ to (0)) \ frac (y) (\ tg (y)) = \ lim_ (y \ to (0)) \ frac (1) (\ frac (\ tg (y)) ( y)) = \ frac (1) (\ displaystyle \ lim_ (y \ to (0)) \ frac (\ tg (y)) (y)) = \ frac (1) (1) = 1. $$

The equality $ \ lim _ (\ alpha \ to (0)) \ frac (\ arctg \ alpha) (\ alpha) = 1 $ is proved.

Equalities a), b), c) are often used along with the first remarkable limit.

Example No. 2

Compute limit of $ \ lim_ (x \ to (2)) \ frac (\ sin \ left (\ frac (x ^ 2-4) (x + 7) \ right)) (\ frac (x ^ 2-4) ( x + 7)) $.

Since $ \ lim_ (x \ to (2)) \ frac (x ^ 2-4) (x + 7) = \ frac (2 ^ 2-4) (2 + 7) = 0 $ and $ \ lim_ ( x \ to (2)) \ sin \ left (\ frac (x ^ 2-4) (x + 7) \ right) = \ sin (0) = 0 $, i.e. both the numerator and denominator of the fraction simultaneously tend to zero, then here we are dealing with an uncertainty of the form $ \ frac (0) (0) $, i.e. done. In addition, it can be seen that the expressions under the sine sign and in the denominator coincide (i.e., and are satisfied):

So, both conditions listed at the beginning of the page are met. From this it follows that the formula is applicable, i.e. $ \ lim_ (x \ to (2)) \ frac (\ sin \ left (\ frac (x ^ 2-4) (x + 7) \ right)) (\ frac (x ^ 2-4) (x + 7)) = 1 $.

Answer: $ \ lim_ (x \ to (2)) \ frac (\ sin \ left (\ frac (x ^ 2-4) (x + 7) \ right)) (\ frac (x ^ 2-4) (x +7)) = 1 $.

Example No. 3

Find $ \ lim_ (x \ to (0)) \ frac (\ sin (9x)) (x) $.

Since $ \ lim_ (x \ to (0)) \ sin (9x) = 0 $ and $ \ lim_ (x \ to (0)) x = 0 $, we are dealing with an uncertainty of the form $ \ frac (0 ) (0) $, i.e. done. However, the expressions under the sine sign and in the denominator do not match. Here you need to fit the expression in the denominator to the desired shape. We need the expression $ 9x $ in the denominator - then it will become true. In fact, we are missing the $ 9 multiplier in the denominator, which is not so difficult to introduce - just multiply the denominator expression by $ 9. Naturally, to compensate for the multiplication by $ 9, you will have to immediately by $ 9 and divide:

$$ \ lim_ (x \ to (0)) \ frac (\ sin (9x)) (x) = \ left | \ frac (0) (0) \ right | = \ lim_ (x \ to (0)) \ frac (\ sin (9x)) (9x \ cdot \ frac (1) (9)) = 9 \ lim_ (x \ to (0)) \ frac (\ sin (9x)) (9x) $$

Now the expressions in the denominator and under the sine sign coincide. Both conditions for the limit $ \ lim_ (x \ to (0)) \ frac (\ sin (9x)) (9x) $ are satisfied. Therefore, $ \ lim_ (x \ to (0)) \ frac (\ sin (9x)) (9x) = 1 $. This means that:

$$ 9 \ lim_ (x \ to (0)) \ frac (\ sin (9x)) (9x) = 9 \ cdot (1) = 9. $$

Answer: $ \ lim_ (x \ to (0)) \ frac (\ sin (9x)) (x) = 9 $.

Example No. 4

Find $ \ lim_ (x \ to (0)) \ frac (\ sin (5x)) (\ tg (8x)) $.

Since $ \ lim_ (x \ to (0)) \ sin (5x) = 0 $ and $ \ lim_ (x \ to (0)) \ tg (8x) = 0 $, here we are dealing with an uncertainty of the form $ \ frac (0) (0) $. However, the shape of the first remarkable limit is violated. A numerator containing $ \ sin (5x) $ requires $ 5x $ in the denominator. In this situation, the easiest way is to divide the numerator by $ 5x $, and then multiply by $ 5x $. In addition, we will perform a similar operation with the denominator, multiplying and dividing $ \ tg (8x) $ by $ 8x $:

$$ \ lim_ (x \ to (0)) \ frac (\ sin (5x)) (\ tg (8x)) = \ left | \ frac (0) (0) \ right | = \ lim_ (x \ to (0)) \ frac (\ frac (\ sin (5x)) (5x) \ cdot (5x)) (\ frac (\ tg (8x)) (8x) \ cdot (8x) ) $$

Reducing by $ x $ and moving the constant $ \ frac (5) (8) $ outside the limit sign, we get:

$$ \ lim_ (x \ to (0)) \ frac (\ frac (\ sin (5x)) (5x) \ cdot (5x)) (\ frac (\ tg (8x)) (8x) \ cdot (8x )) = \ frac (5) (8) \ cdot \ lim_ (x \ to (0)) \ frac (\ frac (\ sin (5x)) (5x)) (\ frac (\ tg (8x)) ( 8x)) $$

Note that $ \ lim_ (x \ to (0)) \ frac (\ sin (5x)) (5x) $ fully meets the requirements for the first remarkable limit. To find $ \ lim_ (x \ to (0)) \ frac (\ tg (8x)) (8x) $ the formula is applicable:

$$ \ frac (5) (8) \ cdot \ lim_ (x \ to (0)) \ frac (\ frac (\ sin (5x)) (5x)) (\ frac (\ tg (8x)) (8x )) = \ frac (5) (8) \ cdot \ frac (\ displaystyle \ lim_ (x \ to (0)) \ frac (\ sin (5x)) (5x)) (\ displaystyle \ lim_ (x \ to (0)) \ frac (\ tg (8x)) (8x)) = \ frac (5) (8) \ cdot \ frac (1) (1) = \ frac (5) (8). $$

Answer: $ \ lim_ (x \ to (0)) \ frac (\ sin (5x)) (\ tg (8x)) = \ frac (5) (8) $.

Example No. 5

Find $ \ lim_ (x \ to (0)) \ frac (\ cos (5x) - \ cos ^ 3 (5x)) (x ^ 2) $.

Since $ \ lim_ (x \ to (0)) (\ cos (5x) - \ cos ^ 3 (5x)) = 1-1 = 0 $ (remember that $ \ cos (0) = 1 $) and $ \ lim_ (x \ to (0)) x ^ 2 = 0 $, then we are dealing with an uncertainty of the form $ \ frac (0) (0) $. However, to apply the first remarkable limit, you need to get rid of the cosine in the numerator by moving to sines (in order to apply the formula later) or tangents (in order to apply the formula later). This can be done with the following transformation:

$$ \ cos (5x) - \ cos ^ 3 (5x) = \ cos (5x) \ cdot \ left (1- \ cos ^ 2 (5x) \ right) $$ $$ \ cos (5x) - \ cos ^ 3 (5x) = \ cos (5x) \ cdot \ left (1- \ cos ^ 2 (5x) \ right) = \ cos (5x) \ cdot \ sin ^ 2 (5x). $$

Let's go back to the limit:

$$ \ lim_ (x \ to (0)) \ frac (\ cos (5x) - \ cos ^ 3 (5x)) (x ^ 2) = \ left | \ frac (0) (0) \ right | = \ lim_ (x \ to (0)) \ frac (\ cos (5x) \ cdot \ sin ^ 2 (5x)) (x ^ 2) = \ lim_ (x \ to (0)) \ left (\ cos (5x) \ cdot \ frac (\ sin ^ 2 (5x)) (x ^ 2) \ right) $$

The fraction $ \ frac (\ sin ^ 2 (5x)) (x ^ 2) $ is already close to the form required for the first remarkable limit. Let's work a little with the fraction $ \ frac (\ sin ^ 2 (5x)) (x ^ 2) $, adjusting it to the first remarkable limit (note that the expressions in the numerator and under the sine must match):

$$ \ frac (\ sin ^ 2 (5x)) (x ^ 2) = \ frac (\ sin ^ 2 (5x)) (25x ^ 2 \ cdot \ frac (1) (25)) = 25 \ cdot \ frac (\ sin ^ 2 (5x)) (25x ^ 2) = 25 \ cdot \ left (\ frac (\ sin (5x)) (5x) \ right) ^ 2 $$

Let's return to the considered limit:

$$ \ lim_ (x \ to (0)) \ left (\ cos (5x) \ cdot \ frac (\ sin ^ 2 (5x)) (x ^ 2) \ right) = \ lim_ (x \ to (0 )) \ left (25 \ cos (5x) \ cdot \ left (\ frac (\ sin (5x)) (5x) \ right) ^ 2 \ right) = \\ = 25 \ cdot \ lim_ (x \ to ( 0)) \ cos (5x) \ cdot \ lim_ (x \ to (0)) \ left (\ frac (\ sin (5x)) (5x) \ right) ^ 2 = 25 \ cdot (1) \ cdot ( 1 ^ 2) = 25. $$

Answer: $ \ lim_ (x \ to (0)) \ frac (\ cos (5x) - \ cos ^ 3 (5x)) (x ^ 2) = 25 $.

Example No. 6

Find the limit $ \ lim_ (x \ to (0)) \ frac (1- \ cos (6x)) (1- \ cos (2x)) $.

Since $ \ lim_ (x \ to (0)) (1- \ cos (6x)) = 0 $ and $ \ lim_ (x \ to (0)) (1- \ cos (2x)) = 0 $, then we are dealing with the uncertainty $ \ frac (0) (0) $. Let's open it up with the first remarkable limit. To do this, let's move from cosines to sines. Since $ 1- \ cos (2 \ alpha) = 2 \ sin ^ 2 (\ alpha) $, then:

$$ 1- \ cos (6x) = 2 \ sin ^ 2 (3x); \; 1- \ cos (2x) = 2 \ sin ^ 2 (x). $$

Passing in the given limit to the sines, we will have:

$$ \ lim_ (x \ to (0)) \ frac (1- \ cos (6x)) (1- \ cos (2x)) = \ left | \ frac (0) (0) \ right | = \ lim_ (x \ to (0)) \ frac (2 \ sin ^ 2 (3x)) (2 \ sin ^ 2 (x)) = \ lim_ (x \ to (0)) \ frac (\ sin ^ 2 (3x)) (\ sin ^ 2 (x)) = \\ = \ lim_ (x \ to (0)) \ frac (\ frac (\ sin ^ 2 (3x)) ((3x) ^ 2) \ cdot (3x) ^ 2) (\ frac (\ sin ^ 2 (x)) (x ^ 2) \ cdot (x ^ 2)) = \ lim_ (x \ to (0)) \ frac (\ left (\ frac (\ sin (3x)) (3x) \ right) ^ 2 \ cdot (9x ^ 2)) (\ left (\ frac (\ sin (x)) (x) \ right) ^ 2 \ cdot (x ^ 2)) = 9 \ cdot \ frac (\ displaystyle \ lim_ (x \ to (0)) \ left (\ frac (\ sin (3x)) (3x) \ right) ^ 2) (\ displaystyle \ lim_ (x \ to (0)) \ left (\ frac (\ sin (x)) (x) \ right) ^ 2) = 9 \ cdot \ frac (1 ^ 2) (1 ^ 2) = 9. $$

Answer: $ \ lim_ (x \ to (0)) \ frac (1- \ cos (6x)) (1- \ cos (2x)) = 9 $.

Example No. 7

Compute the limit of $ \ lim_ (x \ to (0)) \ frac (\ cos (\ alpha (x)) - \ cos (\ beta (x))) (x ^ 2) $ assuming $ \ alpha \ neq \ beta $.

Detailed explanations were given earlier, but here we just note that there is again the uncertainty $ \ frac (0) (0) $. Let's go from cosines to sines using the formula

$$ \ cos \ alpha- \ cos \ beta = -2 \ sin \ frac (\ alpha + \ beta) (2) \ cdot \ sin \ frac (\ alpha- \ beta) (2). $$

Using the above formula, we get:

$$ \ lim_ (x \ to (0)) \ frac (\ cos (\ alpha (x)) - \ cos (\ beta (x))) (x ^ 2) = \ left | \ frac (0) ( 0) \ right | = \ lim_ (x \ to (0)) \ frac (-2 \ sin \ frac (\ alpha (x) + \ beta (x)) (2) \ cdot \ sin \ frac (\ alpha (x) - \ beta (x)) (2)) (x ^ 2) = \\ = -2 \ cdot \ lim_ (x \ to (0)) \ frac (\ sin \ left (x \ cdot \ frac (\ alpha + \ beta ) (2) \ right) \ cdot \ sin \ left (x \ cdot \ frac (\ alpha- \ beta) (2) \ right)) (x ^ 2) = -2 \ cdot \ lim_ (x \ to ( 0)) \ left (\ frac (\ sin \ left (x \ cdot \ frac (\ alpha + \ beta) (2) \ right)) (x) \ cdot \ frac (\ sin \ left (x \ cdot \ frac (\ alpha- \ beta) (2) \ right)) (x) \ right) = \\ = -2 \ cdot \ lim_ (x \ to (0)) \ left (\ frac (\ sin \ left (x \ cdot \ frac (\ alpha + \ beta) (2) \ right)) (x \ cdot \ frac (\ alpha + \ beta) (2)) \ cdot \ frac (\ alpha + \ beta) (2) \ cdot \ frac (\ sin \ left (x \ cdot \ frac (\ alpha- \ beta) (2) \ right)) (x \ cdot \ frac (\ alpha- \ beta) (2)) \ cdot \ frac (\ alpha- \ beta) (2) \ right) = \\ = - \ frac ((\ alpha + \ beta) \ cdot (\ alpha- \ beta)) (2) \ lim_ (x \ to (0)) \ frac (\ sin \ left (x \ cdot \ frac (\ alpha + \ beta) (2) \ right)) (x \ cdot \ frac (\ alpha + \ beta) (2)) \ cdot \ lim_ (x \ to (0)) \ frac (\ sin \ left (x \ cdot \ frac (\ alpha- \ beta) (2) \ right)) (x \ cdot \ frac (\ alpha- \ beta) (2)) = - \ frac (\ alpha ^ 2- \ beta ^ 2) (2) \ cdot (1) \ cdot (1) = \ frac (\ beta ^ 2- \ alpha ^ 2) (2). $$

Answer: $ \ lim_ (x \ to (0)) \ frac (\ cos (\ alpha (x)) - \ cos (\ beta (x))) (x ^ 2) = \ frac (\ beta ^ 2- \ alpha ^ 2) (2) $.

Example No. 8

Find the limit $ \ lim_ (x \ to (0)) \ frac (\ tg (x) - \ sin (x)) (x ^ 3) $.

Since $ \ lim_ (x \ to (0)) (\ tg (x) - \ sin (x)) = 0 $ (remember that $ \ sin (0) = \ tg (0) = 0 $) and $ \ lim_ (x \ to (0)) x ^ 3 = 0 $, then here we are dealing with an uncertainty of the form $ \ frac (0) (0) $. Let's open it as follows:

$$ \ lim_ (x \ to (0)) \ frac (\ tg (x) - \ sin (x)) (x ^ 3) = \ left | \ frac (0) (0) \ right | = \ lim_ (x \ to (0)) \ frac (\ frac (\ sin (x)) (\ cos (x)) - \ sin (x)) (x ^ 3) = \ lim_ (x \ to ( 0)) \ frac (\ sin (x) \ cdot \ left (\ frac (1) (\ cos (x)) - 1 \ right)) (x ^ 3) = \ lim_ (x \ to (0)) \ frac (\ sin (x) \ cdot \ left (1- \ cos (x) \ right)) (x ^ 3 \ cdot \ cos (x)) = \\ = \ lim_ (x \ to (0)) \ frac (\ sin (x) \ cdot (2) \ sin ^ 2 \ frac (x) (2)) (x ^ 3 \ cdot \ cos (x)) = \ frac (1) (2) \ cdot \ lim_ (x \ to (0)) \ left (\ frac (\ sin (x)) (x) \ cdot \ left (\ frac (\ sin \ frac (x) (2)) (\ frac (x) ( 2)) \ right) ^ 2 \ cdot \ frac (1) (\ cos (x)) \ right) = \ frac (1) (2) \ cdot (1) \ cdot (1 ^ 2) \ cdot (1 ) = \ frac (1) (2). $$

Answer: $ \ lim_ (x \ to (0)) \ frac (\ tg (x) - \ sin (x)) (x ^ 3) = \ frac (1) (2) $.

Example No. 9

Find the limit $ \ lim_ (x \ to (3)) \ frac (1- \ cos (x-3)) ((x-3) \ tg \ frac (x-3) (2)) $.

Since $ \ lim_ (x \ to (3)) (1- \ cos (x-3)) = 0 $ and $ \ lim_ (x \ to (3)) (x-3) \ tg \ frac (x -3) (2) = 0 $, then there is an uncertainty of the form $ \ frac (0) (0) $. Before proceeding to its expansion, it is convenient to replace the variable in such a way that the new variable tends to zero (note that the variable $ \ alpha \ to 0 $ in the formulas). The easiest way is to introduce the variable $ t = x-3 $. However, for the sake of convenience of further transformations (this benefit can be seen in the course of the solution below), it is worth making the following replacement: $ t = \ frac (x-3) (2) $. Note that both replacements are applicable in this case, just the second replacement will allow you to work less with fractions. Since $ x \ to (3) $, then $ t \ to (0) $.

$$ \ lim_ (x \ to (3)) \ frac (1- \ cos (x-3)) ((x-3) \ tg \ frac (x-3) (2)) = \ left | \ frac (0) (0) \ right | = \ left | \ begin (aligned) & t = \ frac (x-3) (2); \\ & t \ to (0) \ end (aligned) \ right | = \ lim_ (t \ to (0)) \ frac (1- \ cos (2t)) (2t \ cdot \ tg (t)) = \ lim_ (t \ to (0)) \ frac (2 \ sin ^ 2t) (2t \ cdot \ tg (t)) = \ lim_ (t \ to (0)) \ frac (\ sin ^ 2t) (t \ cdot \ tg (t)) = \\ = \ lim_ (t \ to (0)) \ frac (\ sin ^ 2t) (t \ cdot \ frac (\ sin (t)) (\ cos (t))) = \ lim_ (t \ to (0)) \ frac (\ sin (t) \ cos (t)) (t) = \ lim_ (t \ to (0)) \ left (\ frac (\ sin (t)) (t) \ cdot \ cos (t) \ right) = \ lim_ (t \ to (0)) \ frac (\ sin (t)) (t) \ cdot \ lim_ (t \ to (0)) \ cos (t) = 1 \ cdot (1) = 1. $$

Answer: $ \ lim_ (x \ to (3)) \ frac (1- \ cos (x-3)) ((x-3) \ tg \ frac (x-3) (2)) = 1 $.

Example No. 10

Find the Limit $ \ lim_ (x \ to \ frac (\ pi) (2)) \ frac (1- \ sin (x)) (\ left (\ frac (\ pi) (2) -x \ right) ^ 2 ) $.

Again we are dealing with $ \ frac (0) (0) $ uncertainty. Before proceeding to its expansion, it is convenient to change the variable in such a way that the new variable tends to zero (note that the variable $ \ alpha \ to (0) $ in the formulas). The easiest way is to enter the variable $ t = \ frac (\ pi) (2) -x $. Since $ x \ to \ frac (\ pi) (2) $, then $ t \ to (0) $:

$$ \ lim_ (x \ to \ frac (\ pi) (2)) \ frac (1- \ sin (x)) (\ left (\ frac (\ pi) (2) -x \ right) ^ 2) = \ left | \ frac (0) (0) \ right | = \ left | \ begin (aligned) & t = \ frac (\ pi) (2) -x; \\ & t \ to (0) \ end (aligned) \ right | = \ lim_ (t \ to (0)) \ frac (1- \ sin \ left (\ frac (\ pi) (2) -t \ right)) (t ^ 2) = \ lim_ (t \ to (0 )) \ frac (1- \ cos (t)) (t ^ 2) = \\ = \ lim_ (t \ to (0)) \ frac (2 \ sin ^ 2 \ frac (t) (2)) ( t ^ 2) = 2 \ lim_ (t \ to (0)) \ frac (\ sin ^ 2 \ frac (t) (2)) (t ^ 2) = 2 \ lim_ (t \ to (0)) \ frac (\ sin ^ 2 \ frac (t) (2)) (\ frac (t ^ 2) (4) \ cdot (4)) = \ frac (1) (2) \ cdot \ lim_ (t \ to ( 0)) \ left (\ frac (\ sin \ frac (t) (2)) (\ frac (t) (2)) \ right) ^ 2 = \ frac (1) (2) \ cdot (1 ^ 2 ) = \ frac (1) (2). $$

Answer: $ \ lim_ (x \ to \ frac (\ pi) (2)) \ frac (1- \ sin (x)) (\ left (\ frac (\ pi) (2) -x \ right) ^ 2) = \ frac (1) (2) $.

Example No. 11

Find the limits $ \ lim_ (x \ to \ frac (\ pi) (2)) \ frac (1- \ sin (x)) (\ cos ^ 2x) $, $ \ lim_ (x \ to \ frac (2 \ pi) (3)) \ frac (\ tg (x) + \ sqrt (3)) (2 \ cos (x) +1) $.

In this case, we don't have to use the first wonderful limit. Please note: both the first and the second limits contain only trigonometric functions and numbers. Often, in examples of this kind, it is possible to simplify the expression under the limit sign. In this case, after the above simplification and reduction of some factors, the uncertainty disappears. I gave this example with only one purpose: to show that the presence of trigonometric functions under the limit sign does not necessarily mean the application of the first remarkable limit.

Since $ \ lim_ (x \ to \ frac (\ pi) (2)) (1- \ sin (x)) = 0 $ (remember that $ \ sin \ frac (\ pi) (2) = 1 $ ) and $ \ lim_ (x \ to \ frac (\ pi) (2)) \ cos ^ 2x = 0 $ (remember that $ \ cos \ frac (\ pi) (2) = 0 $), then we have dealing with an uncertainty of the form $ \ frac (0) (0) $. However, this does not mean that we need to use the first remarkable limit. To disclose the uncertainty, it is enough to take into account that $ \ cos ^ 2x = 1- \ sin ^ 2x $:

$$ \ lim_ (x \ to \ frac (\ pi) (2)) \ frac (1- \ sin (x)) (\ cos ^ 2x) = \ left | \ frac (0) (0) \ right | = \ lim_ (x \ to \ frac (\ pi) (2)) \ frac (1- \ sin (x)) (1- \ sin ^ 2x) = \ lim_ (x \ to \ frac (\ pi) ( 2)) \ frac (1- \ sin (x)) ((1- \ sin (x)) (1+ \ sin (x))) = \ lim_ (x \ to \ frac (\ pi) (2) ) \ frac (1) (1+ \ sin (x)) = \ frac (1) (1 + 1) = \ frac (1) (2). $$

There is a similar solution in Demidovich's Reshebnik (No. 475). As for the second limit, as in the previous examples of this section, we have an uncertainty of the form $ \ frac (0) (0) $. Why does it arise? It arises because $ \ tg \ frac (2 \ pi) (3) = - \ sqrt (3) $ and $ 2 \ cos \ frac (2 \ pi) (3) = - 1 $. We use these values ​​to transform expressions in the numerator and denominator. The purpose of our actions: write down the sum in the numerator and denominator in the form of a product. By the way, often within a similar view, it is convenient to change a variable, made in such a way that the new variable tends to zero (see, for example, examples # 9 or # 10 on this page). However, in this example, there is no sense in replacing, although, if desired, it is easy to change the variable $ t = x- \ frac (2 \ pi) (3) $.

$$ \ lim_ (x \ to \ frac (2 \ pi) (3)) \ frac (\ tg (x) + \ sqrt (3)) (2 \ cos (x) +1) = \ lim_ (x \ to \ frac (2 \ pi) (3)) \ frac (\ tg (x) + \ sqrt (3)) (2 \ cdot \ left (\ cos (x) + \ frac (1) (2) \ right )) = \ lim_ (x \ to \ frac (2 \ pi) (3)) \ frac (\ tg (x) - \ tg \ frac (2 \ pi) (3)) (2 \ cdot \ left (\ cos (x) - \ cos \ frac (2 \ pi) (3) \ right)) = \\ = \ lim_ (x \ to \ frac (2 \ pi) (3)) \ frac (\ frac (\ sin \ left (x- \ frac (2 \ pi) (3) \ right)) (\ cos (x) \ cos \ frac (2 \ pi) (3))) (- 4 \ sin \ frac (x + \ frac (2 \ pi) (3)) (2) \ sin \ frac (x- \ frac (2 \ pi) (3)) (2)) = \ lim_ (x \ to \ frac (2 \ pi) (3 )) \ frac (\ sin \ left (x- \ frac (2 \ pi) (3) \ right)) (- 4 \ sin \ frac (x + \ frac (2 \ pi) (3)) (2) \ sin \ frac (x- \ frac (2 \ pi) (3)) (2) \ cos (x) \ cos \ frac (2 \ pi) (3)) = \\ = \ lim_ (x \ to \ frac (2 \ pi) (3)) \ frac (2 \ sin \ frac (x- \ frac (2 \ pi) (3)) (2) \ cos \ frac (x- \ frac (2 \ pi) (3 )) (2)) (- 4 \ sin \ frac (x + \ frac (2 \ pi) (3)) (2) \ sin \ frac (x- \ frac (2 \ pi) (3)) (2) \ cos (x) \ cos \ frac (2 \ pi) (3)) = \ lim_ (x \ to \ frac (2 \ pi) (3)) \ frac (\ cos \ frac (x- \ frac (2 \ pi) (3)) (2)) (- 2 \ sin \ frac (x + \ frac (2 \ pi) (3)) (2) \ cos (x) \ cos \ frac (2 \ pi) (3 )) = \\ = \ frac (1) (- 2 \ cdot \ frac (\ sqrt (3)) (2) \ cdot \ left (- \ frac (1) (2) \ right) \ cdot \ left ( - \ frac (1) (2) \ right)) = - \ frac (4 ) (\ sqrt (3)). $$

As you can see, we didn't have to apply the first wonderful limit. Of course, this can be done if desired (see note below), but it is not necessary.

What would be the solution using the first wonderful limit? show \ hide

Using the first remarkable limit, we get:

$$ \ lim_ (x \ to \ frac (2 \ pi) (3)) \ frac (\ sin \ left (x- \ frac (2 \ pi) (3) \ right)) (- 4 \ sin \ frac (x + \ frac (2 \ pi) (3)) (2) \ sin \ frac (x- \ frac (2 \ pi) (3)) (2) \ cos (x) \ cos \ frac (2 \ pi ) (3)) = \\ = \ lim_ (x \ to \ frac (2 \ pi) (3)) \ left (\ frac (\ sin \ left (x- \ frac (2 \ pi) (3) \ right)) (x- \ frac (2 \ pi) (3)) \ cdot \ frac (1) (\ frac (\ sin \ frac (x- \ frac (2 \ pi) (3)) (2)) (\ frac (x- \ frac (2 \ pi) (3)) (2))) \ cdot \ frac (1) (- 2 \ sin \ frac (x + \ frac (2 \ pi) (3)) ( 2) \ cos (x) \ cos \ frac (2 \ pi) (3)) \ right) = 1 \ cdot (1) \ cdot \ frac (1) (- 2 \ cdot \ frac (\ sqrt (3) ) (2) \ cdot \ left (- \ frac (1) (2) \ right) \ cdot \ left (- \ frac (1) (2) \ right)) = - \ frac (4) (\ sqrt ( 3)). $$

Answer: $ \ lim_ (x \ to \ frac (\ pi) (2)) \ frac (1- \ sin (x)) (\ cos ^ 2x) = \ frac (1) (2) $, $ \ lim_ ( x \ to \ frac (2 \ pi) (3)) \ frac (\ tg (x) + \ sqrt (3)) (2 \ cos (x) +1) = - \ frac (4) (\ sqrt ( 3)) $.