Solution of limits lim. Infinitesimal and infinitely large functions. Limits with type uncertainty and a method for their solution

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 a few detailed examples solution 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 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 bring specific 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, 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.

By the way! For our readers there is now a 10% discount on

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 carefully and you will notice that in the numerator we have quadratic equation. 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.

Solving problems on finding limits When solving problems on finding limits, some limits should be remembered so that each time they are not calculated anew. Combining these known limits, we will use the properties indicated in § 4 to find new limits. For convenience, we present the most common limits: Limits l X -o X 6 lim f(x) = f(a), if f (x) is continuous x a If it is known that the function is continuous, then instead of finding the limit, we calculate the value of the function. Example 1. Find lim (x * -6n: + 8). Since the many-X->2 member function is continuous, then lim (x*-6x4- 8) = 2*-6-2 + 8 = 4. x-+2 x*_2x 4-1 Example 2. Find lim -G. . First, we find the pre-X-+1 x ~rx denominator parts: lim [xr-\-bx)= 12 + 5-1 =6; it is not equal to X-Y1 zero, which means that property 4 of § 4 can be applied, then x™i *" + &* ~~ lim (x2 bx) - 12 + 5-1 ""6 1. Limit of the denominator X X zero, so property 4 of § 4 cannot be applied. Since the numerator is a constant number, and the denominator is [x2x)->-0 at x--1, then the whole fraction increases in absolute value indefinitely, i.e. lim " 1 X-*- - 1 x* + x Example 4 Find lim \-ll*"!" "" The limit of the denominator is equal to zero: lim (xr-6lr + 8) = 2*-6-2 + 8 = 0, therefore X property 4 of § 4 is not applicable. But the limit of the numerator is also equal to zero: lim (x2 - 5d; + 6) \u003d 22 - 5-2-f 6 \u003d 0. So, the limits of the numerator and denominator are simultaneously equal to zero. However, the number 2 is the root of both the numerator and denominator, so the fraction can be reduced by the difference is x-2 (by Bezout's theorem). Indeed, x * -5x + 6 (x-2) (x-3) x-3 x "-6x + 8~ (x-2) (x-4) ~~ x-4 "hence, xr--f- 6 r x-3 -1 1 Example 5. Find lim xn (n is an integer, positive). X co each factor grows indefinitely, then the product also grows indefinitely, i.e. lim xn = oo. x oo Example 6. Find lim xn(n is an integer, positive) X -> - CO Since each factor grows in absolute value, remaining negative, then in the case of an even degree the product will grow indefinitely, remaining positive, i.e., lim *n = + oo (for even n). *-* -co In the case of an odd degree, the absolute value of the product increases, but it remains negative, i.e., lim xn = - oo (for odd n). n -- 00 Example 7. Find lim . x x - * - co * If m> ny then you can write: m = n + kt where k>0. Therefore xm b lim -=- = lim -=-= lim x . yP Yn x -x> A x yu Came to example 6. If ty uTL xm I lim lim lim m. lim -b = 0. X-*oo X* It is recommended to remember the result of this example in the following form: The power function grows the faster, the greater the exponent. $ xv_3xg + 7 i.e. xv, then 3 7_ Example 9. Find lira By performing transformations, we get lira ... ^ = lim X CO + 3 7 3 Since lim -5 \u003d 0, lim -, \u003d 0 , then the limit of the denominator is equal to zero, while the limit of the numerator is 1. Therefore, the whole fraction increases indefinitely, i.e. t. lim Calculate the limit S of the denominator, remembering that the cos*-function is continuous: lira (2 + cos x) = 2 + cozy = 2. Then x->- S lim (l-fsin*) Example 15. Find lim *<*-e>2 and lim e "(X" a) \ We set X-+ ± co X ± CO we press (l: - a) 2 \u003d z; since (x - a)2 always grows non-negatively and indefinitely with x, then as x - ± oo the new variable z - * oc. Therefore, we get u £<*-«)* = X ->± 00 s=lim eg = oo (see remark to §5). r -** co. Similarly, lim e~(X-a)2 = lim e~z=Q, since x ± oo r m - (x-a)r decreases without bound as x -> ± oo (see the remark on §

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:

The expressions under the sine sign and in the denominator simultaneously tend to zero, i.e. there is an uncertainty of the form $\frac(0)(0)$.
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 the desired form. 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: $t=\frac(x-3)(2)$. I note that both substitutions are applicable in this case, just the second substitution 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, if desired, the change of the variable $t=x-\frac(2\pi)(3)$ is easy to implement.

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

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

Theory of limits- one of the sections of mathematical analysis, which one can master, others hardly calculate the limits. The question of finding limits is quite general, since there are dozens of tricks limit solutions various kinds. The same limits can be found both by L'Hopital's rule and without it. It happens that the schedule in a series of infinitesimal functions allows you to quickly get the desired result. There are a set of tricks and tricks that allow you to find the limit of a function of any complexity. In this article, we will try to understand the main types of limits that are most often encountered in practice. We will not give the theory and definition of the limit here, there are many resources on the Internet where this is chewed. Therefore, let's do practical calculations, it is here that you start "I don't know! I don't know how! We weren't taught!"

Calculation of limits by the substitution method

Example 1 Find the limit of a function
Lim((x^2-3*x)/(2*x+5),x=3).
Solution: In theory, examples of this kind are calculated by the usual substitution

The limit is 18/11.
There is nothing complicated and wise within such limits - they substituted the value, calculated, wrote down the limit in response. However, on the basis of such limits, everyone is taught that, first of all, you need to substitute a value into the function. Further, the limits complicate, introduce the concept of infinity, uncertainty, and the like.

Limit with uncertainty of type infinity divided by infinity. Uncertainty disclosure methods

Example 2 Find the limit of a function
Lim((x^2+2x)/(4x^2+3x-4),x=infinity).
Solution: A limit of the form polynomial divided by a polynomial is given, and the variable tends to infinity

A simple substitution of the value to which the variable should find the limits will not help, we get an uncertainty of the form infinity divided by infinity.
Pot theory of limits The algorithm for calculating the limit is to find the largest degree of "x" in the numerator or denominator. Next, the numerator and denominator are simplified on it and the limit of the function is found

Since the value tends to zero when the variable goes to infinity, they are neglected, or written in the final expression as zeros

Immediately from practice, you can get two conclusions that are a hint in the calculations. If the variable tends to infinity and the degree of the numerator is greater than the degree of the denominator, then the limit is equal to infinity. Otherwise, if the polynomial in the denominator is higher order than in the numerator, the limit is zero.
The limit formula can be written as

If we have a function of the form of an ordinary log without fractions, then its limit is equal to infinity

next type limits concerns the behavior of functions near zero.

Example 3 Find the limit of a function
Lim((x^2+3x-5)/(x^2+x+2), x=0).
Solution: Here it is not required to take out the leading multiplier of the polynomial. Exactly the opposite, it is necessary to find the smallest power of the numerator and denominator and calculate the limit

x^2 value; x tend to zero when the variable tends to zero Therefore, they are neglected, thus we get

that the limit is 2.5.

Now you know how to find the limit of a function kind of a polynomial divided by a polynomial if the variable tends to infinity or 0. But this is only a small and easy part of the examples. From next material You will learn how to uncover the uncertainties of the limits of a function.

Limit with uncertainty of type 0/0 and methods for its calculation

Immediately everyone remembers the rule according to which you cannot divide by zero. However, the theory of limits in this context means infinitesimal functions.
Let's look at a few examples to illustrate.

Example 4 Find the limit of a function
Lim((3x^2+10x+7)/(x+1), x=-1).
Solution: When substituting the value of the variable x = -1 into the denominator, we get zero, we get the same in the numerator. So we have uncertainty of the form 0/0.
It is easy to deal with such uncertainty: you need to factorize the polynomial, or rather, select a factor that turns the function into zero.

After decomposition, the limit of the function can be written as

That's the whole technique for calculating the limit of a function. We do the same if there is a limit of the form of a polynomial divided by a polynomial.

Example 5 Find the limit of a function
Lim((2x^2-7x+6)/(3x^2-x-10), x=2).
Solution: Direct substitution shows
2*4-7*2+6=0;
3*4-2-10=0
what do we have type uncertainty 0/0.
Divide the polynomials by the factor that introduces the singularity

There are teachers who teach that polynomials of the 2nd order, that is, the type of "quadratic equations" should be solved through the discriminant. But real practice shows that it is longer and more complicated, so get rid of features within the limits according to the specified algorithm. Thus, we write the function in the form prime factors and count to the limit

As you can see, there is nothing complicated in calculating such limits. You know how to divide polynomials at the time of studying the limits, at least according to the program, you should already pass.
Among the tasks for type uncertainty 0/0 there are those in which it is necessary to apply the formulas of abbreviated multiplication. But if you do not know them, then by dividing the polynomial by the monomial, you can get the desired formula.

Example 6 Find the limit of a function
Lim((x^2-9)/(x-3), x=3).
Solution: We have an uncertainty of type 0/0 . In the numerator, we use the formula for abbreviated multiplication

and calculate the desired limit

Uncertainty disclosure method by multiplication by the conjugate

The method is applied to the limits in which irrational functions generate uncertainty. The numerator or denominator turns to zero at the calculation point and it is not known how to find the boundary.

Example 7 Find the limit of a function
Lim((sqrt(x+2)-sqrt(7x-10))/(3x-6), x=2).
Solution: Let's represent the variable in the limit formula

When substituting, we get an uncertainty of type 0/0.
According to the theory of limits, the scheme for bypassing this singularity consists in multiplying an irrational expression by its conjugate. To keep the expression unchanged, the denominator must be divided by the same value

By the difference of squares rule, we simplify the numerator and calculate the limit of the function

We simplify the terms that create a singularity in the limit and perform the substitution

Example 8 Find the limit of a function
Lim((sqrt(x-2)-sqrt(2x-5))/(3-x), x=3).
Solution: Direct substitution shows that the limit has a singularity of the form 0/0.

To expand, multiply and divide by the conjugate to the numerator

Write down the difference of squares

We simplify the terms that introduce a singularity and find the limit of the function

Example 9 Find the limit of a function
Lim((x^2+x-6)/(sqrt(3x-2)-2), x=2).
Solution: Substitute the deuce in the formula

Get uncertainty 0/0.
The denominator must be multiplied by the conjugate expression, and in the numerator, solve the quadratic equation or factorize, taking into account the singularity. Since it is known that 2 is a root, then the second root is found by the Vieta theorem

Thus, we write the numerator in the form

and put in the limit

Having reduced the difference of squares, we get rid of the features in the numerator and denominator

In the above way, you can get rid of the singularity in many examples, and the application should be noticed everywhere where the given difference of the roots turns into zero when substituting. Other types of limits concern exponential functions, infinitesimal functions, logarithms, singular limits, and other techniques. But you can read about this in the articles below on limits.

When calculating limits, consider the following ground 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 factors:

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

4. A constant factor can be taken out of the limit sign:

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

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

Finding the limit of a function should begin with substituting the value into the expression for the function. However, if it turns out numerical value 0 or ¥, then the desired limit is found.

Example 2.1. Calculate 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 in such a way as to reveal this uncertainty.

An indeterminacy of the 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 factorize the polynomials and reduce by a common factor. This factor is zero for limit value X .

Example 2.2. Calculate limit.

Solution.

Substituting , we get the uncertainty:

Let's factorize the numerator and denominator:

;

We reduce by a common factor and get

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

Example 2.3. Calculate limit.

Solution. Substituting ∞ results in an uncertainty of the form , so we divide all terms of the expression by x 3.

Here 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 the adjoint expression.

Example 2.4. Calculate Limit

Solution.

When calculating limits for disclosure of 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.

Consider the example of Ya. I. Perelman, which gives 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 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-bearing money is added to the fixed capital only after a year, then by this time 100 den. units will turn into 200 den.

Now let's see what 100 den will turn into. units, if interest money is added to the fixed capital every six months. After half a year 100 den. units will grow by 100 × 1.5 = 150, and in another six months - by 150 × 1.5 = 225 (money 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 timeframe for adding interest money to 0.1 year, 0.01 year, 0.001 year, and so on. Then out of 100 den. units a year later:

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 of joining interest, the accrued capital does not grow indefinitely, but approaches a certain limit equal to approximately 271. The capital placed at 100% per annum cannot increase more than 2.71 times, even if the accrued interest were added to the capital every second, because

Example 2.5. Calculate function limit

Solution.

Example 2.6. Calculate function limit .

Solution. Substituting we get the uncertainty:

Using 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 function limit

Solution.

To reveal the uncertainty of the form or, you can use the L'Hopital 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 multiple 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

Function continuity

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:

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

Figure 2.3 - Increment of variables

From the definition of a function that is continuous at the point , it follows that . This equality means the fulfillment of three conditions:

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

Consequently, , means - break point

Function derivative