Definition of sequence and function limits, properties of limits, first and second remarkable limits, examples.

constant number a called limit sequences(x n ) if for any arbitrarily small positive number ε > 0 there exists a number N such that all values x n, for which n>N, satisfy the inequality

Write it as follows: or x n → a.

Inequality (6.1) is equivalent to the double inequality

a - ε< x n < a + ε которое означает, что точки x n, starting from some number n>N, lie inside the interval (a-ε , a+ε), i.e. fall into any small ε-neighborhood of the point a.

A sequence that has a limit is called converging, otherwise - divergent.

The concept of the limit of a function is a generalization of the concept of the limit of a sequence, since the limit of a sequence can be considered as the limit of the function x n = f(n) of an integer argument n.

Let a function f(x) be given and let a - limit point the domain of definition of this function D(f), i.e. such a point, any neighborhood of which contains points of the set D(f) different from a. Dot a may or may not belong to the set D(f).

Definition 1. The constant number A is called limit functions f(x) at x→ a if for any sequence (x n ) of argument values ​​tending to a, the corresponding sequences (f(x n)) have the same limit A.

This definition is called defining the limit of a function according to Heine, or " in the language of sequences”.

Definition 2. The constant number A is called limit functions f(x) at x→a if, given an arbitrary, arbitrarily small positive number ε, one can find δ >0 (depending on ε) such that for all x, lying in the ε-neighborhood of the number a, i.e. for x satisfying the inequality
0 < x-a < ε , значения функции f(x) будут лежать в ε-окрестности числа А, т.е. |f(x)-A| < ε

This definition is called defining the limit of a function according to Cauchy, or “in the language ε - δ"

Definitions 1 and 2 are equivalent. If the function f(x) as x → a has limit equal to A, this is written as

In the event that the sequence (f(x n)) increases (or decreases) indefinitely for any method of approximation x to your limit a, then we will say that the function f(x) has infinite limit, and write it as:

A variable (i.e. a sequence or function) whose limit is zero is called infinitely small.

A variable whose limit is equal to infinity is called infinitely large.

To find the limit in practice, use the following theorems.

Theorem 1 . If every limit exists

(6.4)

(6.5)

(6.6)

Comment. Expressions of the form 0/0, ∞/∞, ∞-∞ 0*∞ are indefinite, for example, the ratio of two infinitesimal or infinitely large quantities, and finding a limit of this kind is called “uncertainty disclosure”.

Theorem 2.

those. it is possible to pass to the limit at the base of the degree at a constant exponent, in particular,

Theorem 3.

(6.11)

where e» 2.7 is the base of the natural logarithm. Formulas (6.10) and (6.11) are called the first remarkable limit and the second remarkable limit.

The corollaries of formula (6.11) are also used in practice:

(6.12)

(6.13)

(6.14)

in particular the limit

If x → a and at the same time x > a, then write x →a + 0. If, in particular, a = 0, then write +0 instead of the symbol 0+0. Similarly, if x→a and at the same time x and are named accordingly. right limit and left limit functions f(x) at the point a. For the limit of the function f(x) to exist as x→ a, it is necessary and sufficient that . The function f(x) is called continuous at the point x 0 if limit

(6.15)

Condition (6.15) can be rewritten as:

that is, passage to the limit under the sign of a function is possible if it is continuous at a given point.

If equality (6.15) is violated, then we say that at x = xo function f(x) It has gap. Consider the function y = 1/x. The domain of this function is the set R, except for x = 0. The point x = 0 is a limit point of the set D(f), since in any of its neighborhoods, i.e., any open interval containing the point 0 contains points from D(f), but it does not itself belong to this set. The value f(x o)= f(0) is not defined, so the function has a discontinuity at the point x o = 0.

The function f(x) is called continuous on the right at a point x o if limit

and continuous on the left at a point x o if limit

Continuity of a function at a point x o is equivalent to its continuity at this point both on the right and on the left.

For a function to be continuous at a point x o, for example, on the right, it is necessary, firstly, that there is a finite limit , and secondly, that this limit be equal to f(x o). Therefore, if at least one of these two conditions is not met, then the function will have a gap.

1. If the limit exists and is not equal to f(x o), then they say that function f(x) at the point xo has break of the first kind, or jump.

2. If the limit is +∞ or -∞ or does not exist, then they say that in point x o the function has a break second kind.

For example, the function y = ctg x as x → +0 has a limit equal to +∞ , which means that at the point x=0 it has a discontinuity of the second kind. Function y = E(x) (integer part of x) at points with integer abscissas has discontinuities of the first kind, or jumps.

A function that is continuous at every point of the interval is called continuous in . A continuous function is represented by a solid curve.

Many problems associated with the continuous growth of some quantity lead to the second remarkable limit. Such tasks, for example, include: the growth of the contribution according to the law of compound interest, the growth of the country's population, the decay of a radioactive substance, the multiplication of bacteria, etc.

Consider example of Ya. I. Perelman, which gives the interpretation of the number e in the compound interest problem. Number e there is a limit . 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 the limit

Example 3.1. Using the definition of the limit of a number sequence, prove that the sequence x n =(n-1)/n has a limit equal to 1.

Solution. We need to prove that whatever ε > 0 we take, there is a natural number N for it, such that for all n > N the inequality |x n -1|< ε

Take any ε > 0. Since x n -1 =(n+1)/n - 1= 1/n, then to find N it is enough to solve the inequality 1/n<ε. Отсюда n>1/ε and, therefore, N can be taken as the integer part of 1/ε N = E(1/ε). We thus proved that the limit .

Example 3.2. Find the limit of a sequence given by a common term .

Solution. Apply the limit sum theorem and find the limit of each term. As n → ∞, the numerator and denominator of each term tends to infinity, and we cannot apply the quotient limit theorem directly. Therefore, we first transform x n, dividing the numerator and denominator of the first term by n 2, and the second n. Then, applying the quotient limit theorem and the sum limit theorem, we find:

Example 3.3. . Find .

Solution.

Here we have used the degree limit theorem: the limit of a degree is equal to the degree of the limit of the base.

Example 3.4. Find ( ).

Solution. It is impossible to apply the difference limit theorem, since we have an uncertainty of the form ∞-∞. Let's transform the formula of the general term:

Example 3.5. Given a function f(x)=2 1/x . Prove that the limit does not exist.

Solution. We use the definition 1 of the limit of a function in terms of a sequence. Take a sequence ( x n ) converging to 0, i.e. Let us show that the value f(x n)= behaves differently for different sequences. Let x n = 1/n. Obviously, then the limit Let's choose now as x n a sequence with a common term x n = -1/n, also tending to zero. Therefore, there is no limit.

Example 3.6. Prove that the limit does not exist.

Solution. Let x 1 , x 2 ,..., x n ,... be a sequence for which
. How does the sequence (f(x n)) = (sin x n ) behave for different x n → ∞

If x n \u003d p n, then sin x n \u003d sin (p n) = 0 for all n and limit If
xn=2 p n+ p /2, then sin x n = sin(2 p n+ p /2) = sin p /2 = 1 for all n and hence the limit. Thus does not exist.

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 of the 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.

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

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Methods for solving limits. Uncertainties.
Function growth order. Replacement Method

Example 4

Find the limit

This is a simpler example for a do-it-yourself solution. In the proposed example, again, uncertainty (of a higher order of growth than the root).

If "x" tends to "minus infinity"

The ghost of "minus infinity" has long been hovering in this article. Consider limits with polynomials in which . The principles and methods of solution will be exactly the same as in the first part of the lesson, with the exception of a number of nuances.

Consider 4 chips that will be required to solve practical tasks:

1) Calculate the limit

The value of the limit depends only on the term because it has the highest order of growth. If , then infinitely large modulo negative number to the power of EVEN, in this case - in the fourth, is equal to "plus infinity": . Constant ("two") positive, that's why:

2) Calculate the limit

Here is the senior degree again even, that's why: . But there is a "minus" in front ( negative constant –1), therefore:

3) Calculate the limit

The value of the limit depends only on . As you remember from school, "minus" "pops out" from under the odd degree, so infinitely large modulo negative number to an ODD power equals "minus infinity", in this case: .
Constant ("four") positive, means:

4) Calculate the limit

The first guy in the village has again odd degree, moreover, in the bosom negative constant, which means: Thus:
.

Example 5

Find the limit

Using the above points, we conclude that there is uncertainty here. The numerator and denominator are of the same order of growth, which means that in the limit a finite number will be obtained. We learn the answer by discarding all the fry:

The solution is trivial:

Example 6

Find the limit

This is a do-it-yourself example. Full solution and answer at the end of the lesson.

And now, perhaps the most subtle of the cases:

Example 7

Find the limit

Considering the senior terms, we come to the conclusion that there is uncertainty here. The numerator is of a higher order of growth than the denominator, so we can immediately say that the limit is infinity. But what kind of infinity, "plus" or "minus"? The reception is the same - in the numerator and denominator we will get rid of the little things:

We decide:

Divide the numerator and denominator by

Example 15

Find the limit

This is a do-it-yourself example. An approximate sample of finishing at the end of the lesson.

A couple more interesting examples on the topic of variable substitution:

Example 16

Find the limit

Substituting one into the limit results in uncertainty. The replacement of the variable is already suggesting, but first we convert the tangent using the formula. Indeed, why do we need a tangent?

Note that , therefore . If it is not entirely clear, look at the sine values ​​\u200b\u200bin trigonometric table. Thus, we immediately get rid of the factor , in addition, we get the more familiar uncertainty 0:0. It would be nice if our limit also tended to zero.

Let's replace:

If , then

Under the cosine we have "x", which also needs to be expressed through "te".
From the replacement we express: .

We complete the solution:

(1) Performing the substitution

(2) Expand the brackets under the cosine.

(4) To organize first wonderful limit, artificially multiply the numerator by and the reciprocal of .

Task for independent solution:

Example 17

Find the limit

Full solution and answer at the end of the lesson.

These were simple tasks in their class; in practice, everything is worse, and, in addition to reduction formulas, one has to use different trigonometric formulas, as well as other tricks. In the article Complex Limits, I analyzed a couple of real examples =)

On the eve of the holiday, we will finally clarify the situation with one more common uncertainty:

Elimination of uncertainty "one to the power of infinity"

This uncertainty is “served” second wonderful limit, and in the second part of that lesson, we looked in great detail at standard examples of solutions that are found in practice in most cases. Now the picture with exhibitors will be completed, in addition, the final tasks of the lesson will be devoted to the limits-"tricks" in which it seems that it is necessary to apply the 2nd wonderful limit, although this is not at all the case.

The disadvantage of the two working formulas of the 2nd remarkable limit is that the argument must tend to "plus infinity" or to zero. But what if the argument tends to a different number?

The universal formula comes to the rescue (which is actually a consequence of the second remarkable limit):

Uncertainty can be eliminated by the formula:

Somewhere like I already explained what the square brackets mean. Nothing special, brackets are just brackets. Usually they are used to clearly highlight a mathematical notation.

Let's highlight the essential points of the formula:

1) It's about only about uncertainty and no other.

2) Argument "x" can tend to arbitrary value(and not only to zero or ), in particular, to "minus infinity" or to anyone final number.

Using this formula, you can solve all the examples of the lesson Remarkable Limits, which belong to the 2nd wonderful limit. For example, let's calculate the limit:

In this case , and according to the formula :

True, I don’t advise you to do this, in the tradition, you still use the “usual” design of the solution, if it can be applied. However using the formula is very convenient to check"classical" examples to the 2nd wonderful limit.