The first wonderful limit. Calculating the limits of functions online Online solution 1 of the first remarkable limit

For those who want to learn how to find the limits in this article we will talk about it. We will not delve into the theory, it is usually given in lectures by teachers. 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 the limit is quite important in the study of the course of higher mathematics, especially when you come across the integral calculus and understand the relationship between the limit and the integral. In the current material, simple examples will be considered, as well as ways to solve them.

Solution examples

What to do with the uncertainty of the form: $ \bigg [\frac(0)(0) \bigg ] $

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

Algorithm for calculating limits

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

1. Substitute point x in the expression following the limit sign. If a certain number is obtained, or infinity, then the limit is completely solved. Otherwise, we have uncertainty: "zero divided by zero" or "infinity divided by infinity" and proceed to the next paragraphs of the instruction.
2. To eliminate the uncertainty "zero divide by zero" you need to factorize the numerator and denominator. Reduce similar. Substitute the 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 x of the greatest degree. We shorten the x's. We substitute x values ​​from under the limit into the remaining expression.

In this article, you got acquainted with the basics of solving limits, often used in the Calculus course. Of course, these are not all types of problems offered by examiners, but only the simplest limits. We will talk about other types of tasks in future articles, 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'Hopital's rule.

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

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Limits give all students of mathematics a lot of trouble. To solve the limit, sometimes you have to use a lot of tricks and choose from a variety of solutions exactly the one that is suitable for a particular example.

In this article, we will not help you understand the limits of your abilities 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 some detailed examples of solving limits with explanations.

The concept of a limit in mathematics

The first question is: what is the limit and the limit of what? 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 indefinitely approaches a certain number a , then a is the limit of this value.

For a function defined in some interval f(x)=y the limit is the number A , to which the function tends when X tending to a certain point a . Dot a belongs to the interval on which the function is defined.

It sounds cumbersome, but it is written very simply:

Lim- from English limit- 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 X tends to some value, this means that the variable does not take on the value of a number, but approaches it infinitely close.

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

To solve this example, we 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 X can tend to any value. It can be any number or infinity. Here is an example when X tends to infinity:

It is intuitively clear that the larger the number in the denominator, the smaller the value will be taken by the function. So, with unlimited growth X 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 X . However, this is the simplest case. Often finding the limit is not so obvious. Within the limits there are uncertainties of type 0/0 or infinity/infinity . What to do in such cases? Use tricks!

Uncertainties within

Uncertainty of the form infinity/infinity

Let there be a limit:

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

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

To uncover type ambiguities infinity/infinity divide the numerator and denominator by X to the highest degree.

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Another type of uncertainty: 0/0

As always, substitution into 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. Let's find the roots and write:

Let's reduce and get:

So, if you encounter type ambiguity 0/0 - factorize the numerator and denominator.

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

L'Hopital's rule within

Another powerful way to eliminate both types of uncertainties. 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.

Visually, L'Hopital's rule looks like this:

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

And now a real example:

There is a typical uncertainty 0/0 . Take the derivatives of the numerator and denominator:

Voila, the uncertainty is eliminated quickly and elegantly.

We hope that you will be able to put this information to good use in practice and find the answer to the question "how to solve limits in higher mathematics". If you need to calculate the limit of a sequence or the limit of a function at a point, and there is no time for this work from the word “absolutely”, contact a professional student service for a quick and detailed solution.

Function limit- number a will be the limit of some variable value if, in the process of its change, this variable approaches indefinitely a.

Or in other words, the number A is the limit of the function y=f(x) at the point x0, if for any sequence of points from the domain of definition of the function , not equal to x0, and which converges to the point x 0 (lim x n = x0), the sequence of corresponding values ​​of the function converges to the number A.

Graph of a function whose limit with an argument that tends to infinity is L:

Meaning BUT is limit (limit value) of the function f(x) at the point x0 if for any sequence of points , which converges to x0, but which does not contain x0 as one of its elements (i.e. in the punctured neighborhood x0), the sequence of function values converges to A.

The limit of a function according to Cauchy.

Meaning A will be function limit f(x) at the point x0 if for any forward taken non-negative number ε a non-negative corresponding number will be found δ = δ(ε) such that for each argument x, satisfying the condition 0 < | x - x0 | < δ , the inequality | f(x) A |< ε .

It will be very simple if you understand the essence of the limit and the basic rules for finding it. That the limit of the function f(x) at x aspiring to a equals A, is written like this:

Moreover, the value to which the variable tends x, can be not only a number, but also infinity (∞), sometimes +∞ or -∞, or there may be no limit at all.

To understand how find the limits of a function, it is best to see examples of solutions.

We need to find the limits of the function f(x) = 1/x at:

x→ 2, x→ 0, x→ ∞.

Let's find the solution of the first limit. To do this, you can simply substitute x the number to which it aspires, i.e. 2, we get:

Find the second limit of the function. Here, substitute in pure form 0 instead of x it is impossible, because cannot be divided by 0. But we can take values ​​close to zero, for example, 0.01; 0.001; 0.0001; 0.00001 and so on, with the value of the function f(x) will increase: 100; 1000; 10000; 100000 and so on. Thus, it can be understood that when x→ 0 the value of the function that is under the limit sign will increase indefinitely, i.e. strive for infinity. Which means:

Regarding the third limit. The same situation as in the previous case, it is impossible to substitute ∞ in its purest form. We need to consider the case of unlimited increase x. We alternately substitute 1000; 10000; 100000 and so on, we have that the value of the function f(x) = 1/x will decrease: 0.001; 0.0001; 0.00001; and so on, tending to zero. That's why:

It is necessary to calculate the limit of the function

Starting to solve the second example, we see the uncertainty. From here we find the highest degree of the numerator and denominator - this is x 3, we take it out of brackets in the numerator and denominator and then reduce it by it:

Answer

The first step in finding this limit, substitute the value 1 instead of x, resulting in the uncertainty . To solve it, we decompose the numerator into factors , we will do this by finding the roots quadratic equation x 2 + 2x - 3:

D \u003d 2 2 - 4 * 1 * (-3) \u003d 4 +12 \u003d 16→ √ D=√16 = 4

x 1,2 = (-2± 4) / 2→ x 1 \u003d -3;x2= 1.

So the numerator would be:

Answer

This is the definition of its specific value or a specific area where the function falls, which is limited by the limit.

To decide the limits, follow the rules:

Having understood the essence and main limit decision rules, you will get a basic understanding of how to solve them.

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 $\sin\alpha\to(0)$, we say that the first remarkable limit reveals an indeterminacy 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. The 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.

Corollaries 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 are 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 little or no comment, as detailed explanations were given in the previous examples. When solving, some trigonometric formulas are used, which can be found.

Note that the presence trigonometric functions coupled with the uncertainty of $\frac (0) (0)$ 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 replacement $\alpha=\sin(y)$. Since $\sin(0)=0$, then from the condition $\alpha\to(0)$ we have $y\to(0)$. In addition, there is a neighborhood of zero where $\arcsin\alpha=\arcsin(\sin(y))=y$, 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 replacement $\alpha=\tg(y)$. Since $\tg(0)=0$, the conditions $\alpha\to(0)$ and $y\to(0)$ are equivalent. In addition, there is a neighborhood of zero where $\arctg\alpha=\arctg\tg(y))=y$, therefore, relying on the results of point 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 #2

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

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. and 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. performed. In addition, it can be seen that the expressions under the sine sign and in the denominator are the same (i.e., and is satisfied):

So, both conditions listed at the beginning of the page are met. It follows from this 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 #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., performed. However, the expressions under the sine sign and in the denominator do not match. Here it is required to adjust the expression in the denominator to desired shape. We need the expression $9x$ to be in the denominator - then it will become true. Basically, we're missing the $9$ factor in the denominator, which isn't that hard to enter, just multiply the expression in the denominator by $9$. Naturally, to compensate for the multiplication by $9$, you will have to immediately divide 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 are the same. Both conditions for the limit $\lim_(x\to(0))\frac(\sin(9x))(9x)$ are satisfied. Hence $\lim_(x\to(0))\frac(\sin(9x))(9x)=1$. And 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 #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 indeterminacy of the form $\frac(0)(0)$. However, the form of the first remarkable limit is broken. A numerator containing $\sin(5x)$ requires $5x$ in the denominator. In this situation, the easiest way is to divide the numerator by $5x$, and immediately 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 taking the constant $\frac(5)(8)$ out of 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 satisfies the requirements for the first remarkable limit. To find $\lim_(x\to(0))\frac(\tg(8x))(8x)$ the following 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 #5

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

Since $\lim_(x\to(0))(\cos(5x)-\cos^3(5x))=1-1=0$ (recall that $\cos(0)=1$) and $\lim_(x\to(0))x^2=0$, then we are dealing with an indeterminacy of the form $\frac(0)(0)$. However, in order to apply the first wonderful limit, you should get rid of the cosine in the numerator by going to sines (in order to then apply the formula) or tangents (in order to then apply the formula). You can do this 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 wonderful 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 #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 of $\frac(0)(0)$. Let's open it with the help of 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 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 #7

Calculate limit $\lim_(x\to(0))\frac(\cos(\alpha(x))-\cos(\beta(x)))(x^2)$ given $\alpha\neq\ beta$.

Detailed explanations were given earlier, but here we simply note that again there is an indeterminacy of $\frac(0)(0)$. Let's move 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 #8

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

Since $\lim_(x\to(0))(\tg(x)-\sin(x))=0$ (recall that $\sin(0)=\tg(0)=0$) and $\lim_(x\to(0))x^3=0$, then here we are dealing with an indeterminacy of the form $\frac(0)(0)$. Let's break it down like this:

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

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

Example #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 indeterminacy of the form $\frac(0)(0)$. 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 introduce the variable $t=x-3$. However, for the 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 substitutions 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 #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 the uncertainty of $\frac(0)(0)$. Before proceeding to its expansion, it is convenient to make a variable change in such a way that the new variable tends to zero (note that in the formulas the variable is $\alpha\to(0)$). The easiest way is to introduce 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 #11

Find 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 do not have to use the first wonderful limit. Please note: in both the first and second limits, there are only trigonometric functions and numbers. Often, in examples of this kind, it is possible to simplify the expression located under the limit sign. In this case, after the mentioned 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 $\lim_(x\to\frac(\pi)(2))(1-\sin(x))=0$ (recall that $\sin\frac(\pi)(2)=1$ ) and $\lim_(x\to\frac(\pi)(2))\cos^2x=0$ (recall that $\cos\frac(\pi)(2)=0$), then we have dealing with uncertainty of the form $\frac(0)(0)$. However, this does not mean at all that we need to use the first remarkable limit. To reveal the uncertainty, it suffices 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 solution book (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 the sum in the numerator and denominator as a product. By the way, it is often convenient to change a variable within a similar form so that the new variable tends to zero (see, for example, examples No. 9 or No. 10 on this page). However, in this example, there is no point in replacing the variable, although it is easy to implement the replacement of the variable $t=x-\frac(2\pi)(3)$ if desired.

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

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