Help in solving limits. Examples of finding the limits of functions. Finite limit of a function at infinity

Cauchy definitions of finite and infinite limits of a function at infinity. Definitions of two-sided and one-sided limits (left and right). Examples of solutions to problems in which, using the Cauchy definition, it is required to show that the limit at infinity is equal to a given value, .

Content

See also: Neighborhood of a point
Universal definition of the limit of a function according to Heine and Cauchy

Finite limit of a function at infinity

Function limit at infinity:
|f(x) - a|< ε при |x| >N

Definition of the Cauchy limit
The number a is called the limit of the function f (x) as x tends to infinity (), if
1) there is such |x| >
2) for any arbitrarily small positive number ε > 0 , there exists a number N ε > K, depending on ε , such that for all x, |x| > N ε , the values of the function belong to the ε neighborhood of the point a :
|f (x) - a|< ε .
The limit of a function at infinity is denoted as follows:
.
Or at .

The following notation is also often used:
.

We write this definition using the logical symbols of existence and universality:
.
Here it is assumed that the values belong to the scope of the function.

One-sided limits

Left limit of the function at infinity:
|f(x) - a|< ε при x < -N

Often there are cases when the function is defined only for positive or negative values of the variable x (more precisely, in the vicinity of the point or ). Also, the limits at infinity for positive and negative values of x can have different values. Then one-sided limits are used.

Left limit at infinity or the limit as x tends to minus infinity () is defined as follows:
.
Right limit at infinity or limit as x tends to plus infinity () :
.
One-sided limits at infinity are often written like this:
; .

Infinite function limit at infinity

Infinite function limit at infinity:
|f(x)| > M for |x| > N

Definition of the infinite limit according to Cauchy
Limit of function f (x) when x tends to infinity (), is equal to infinity, if
1) there is such a neighborhood of the point at infinity |x| > K , on which the function is defined (here K is a positive number);
2) for any arbitrarily large number M > 0 , there exists a number N M > K, depending on M , such that for all x, |x| > N M , the values of the function belong to the neighborhood of the point at infinity:
|f (x) | >M.
The infinite limit as x tends to infinity is denoted as follows:
.
Or at .

Using the logical symbols of existence and universality, the definition of the infinite limit of a function can be written as follows:
.

The definitions of the infinite limits of certain signs equal to and are introduced similarly:
.
.

Definitions of one-sided limits at infinity.
Left limits.
.
.
.
Right limits.
.
.
.

Definition of the limit of a function according to Heine

The number a (finite or at infinity) is called the limit of the function f (x) at point x 0 :
,
if
1) there is such a neighborhood of the point at infinity x 0 , on which the function is defined (here or or );
2) for any sequence ( x n ), converging to x 0 : ,
whose elements belong to the neighborhood , the sequence (f(xn)) converges to a :
.

If we take the neighborhood of an unsigned point at infinity as a neighborhood: , then we obtain the definition of the limit of the function as x tends to infinity, . If we take the left-hand or right-hand neighborhood of the point at infinity x 0 : or , then we get the definition of the limit as x tends to minus infinity and plus infinity, respectively.

The Heine and Cauchy definitions of the limit are equivalent.

Examples

Example 1

Using the Cauchy definition, show that
.

Let us introduce the notation:
.
Find the domain of the function . Since the numerator and denominator of a fraction are polynomials, the function is defined for all x except for the points where the denominator vanishes. Let's find these points. We solve a quadratic equation. ;
.
Equation roots:
; .
Since , then and .
Therefore, the function is defined for . This we will use in the future.

We write out the definition of the finite limit of a function at infinity according to Cauchy:
.
Let's transform the difference:
.
Divide the numerator and denominator by and multiply by -1 :
.

Let be .
Then
;
;
;
.

So, we have found that at ,
.
.
Hence it follows that
at , and .

Since it is always possible to increase, we take . Then for any ,
at .
It means that .

Example 2

Let be .
Using the definition of the Cauchy limit, show that:
1) ;
2) .

1) Solution for x tending to minus infinity

Since , then the function is defined for all x .
Let us write out the definition of the limit of the function at equal to minus infinity:
.

Let be . Then
;
.

So, we have found that at ,
.
We enter positive numbers and:
.
It follows that for any positive number M , there is a number , so that for ,
.

It means that .

2) Solution for x tending to plus infinity

Let's transform the original function. Multiply the numerator and denominator of the fraction and apply the difference of squares formula:
.
We have:

.
Let us write the definition of the right limit of the function for :
.

Let's introduce the notation: .
Let's transform the difference:
.
Multiply the numerator and denominator by:
.

Let be
.
Then
;
.

So, we have found that at ,
.
We enter positive numbers and:
.
Hence it follows that
at and .

Since this holds for any positive number, then
.

References:
CM. Nikolsky. Course of mathematical analysis. Volume 1. Moscow, 1983.

Solution examples

Example 1
Calculate a) $ \lim_(x \to 0) \frac(1)(x) $; b)$ \lim_(x \to \infty) \frac(1)(x) $
Decision
a) $$ \lim \limits_(x \to 0) \frac(1)(x) = \infty $$ b)$$ \lim_(x \to \infty) \frac(1)(x) = 0 $$ We often get these limits sent to us asking for help to solve. We decided to highlight them as a separate example and explain that these limits simply need to be remembered, as a rule. If you cannot solve your problem, then send it to us. We will provide a detailed solution. You will be able to familiarize yourself with the progress of the calculation and gather information. This will help you get a credit from the teacher in a timely manner!
Answer
$$ \text(a)) \lim \limits_(x \to \to 0) \frac(1)(x) = \infty \text( b))\lim \limits_(x \to \infty) \frac(1 )(x) = 0 $$

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

Example 3
Solve $ \lim \limits_(x \to -1) \frac(x^2-1)(x+1) $
Decision
As always, we start by substituting the value of $ x $ into the expression under the limit sign. $$ \lim \limits_(x \to -1) \frac(x^2-1)(x+1) = \frac((-1)^2-1)(-1+1)=\frac( 0)(0) $$ What's next? What should be the result? Since this is an uncertainty, this is not yet an answer and we continue the calculation. Since we have a polynomial in the numerators, we decompose it into factors using the familiar formula $$ a^2-b^2=(a-b)(a+b) $$. Remembered? Fine! Now go ahead and apply it with the song :) We get that the numerator $ x^2-1=(x-1)(x+1) $ We continue to solve given the above transformation: $$ \lim \limits_(x \to -1)\frac(x^2-1)(x+1) = \lim \limits_(x \to -1)\frac((x-1)(x+ 1))(x+1) = $$ $$ = \lim \limits_(x \to -1)(x-1)=-1-1=-2 $$
Answer
$$ \lim \limits_(x \to -1) \frac(x^2-1)(x+1) = -2 $$

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

Example 5
Calculate $ \lim \limits_(x \to \infty) \frac(x^2-1)(x+1) $
Decision
$ \lim \limits_(x \to \infty) \frac(x^2-1)(x+1) = \frac(\infty)(\infty) $ What to do? How to be? Do not panic, because the impossible is possible. It is necessary to take out the brackets in both the numerator and the denominator X, and then reduce it. After that, try to calculate the limit. Trying... $$ \lim \limits_(x \to \infty) \frac(x^2-1)(x+1) =\lim \limits_(x \to \infty) \frac(x^2(1-\frac (1)(x^2)))(x(1+\frac(1)(x))) = $$ $$ = \lim \limits_(x \to \infty) \frac(x(1-\frac(1)(x^2)))((1+\frac(1)(x))) = $$ Using the definition from Example 2 and substituting infinity for x, we get: $$ = \frac(\infty(1-\frac(1)(\infty)))((1+\frac(1)(\infty))) = \frac(\infty \cdot 1)(1+ 0) = \frac(\infty)(1) = \infty $$
Answer
$$ \lim \limits_(x \to \infty) \frac(x^2-1)(x+1) = \infty $$

Algorithm for calculating limits

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

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.
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.
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!

When solving problems to find 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 many-X->2

term-function is continuous, then lim (x*-6x4-8) = 2*-6-2 + 8 = 4. x-+2 x*_2x 4-1 Example 2. Find lim -r. . 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. The limit of the denominator X X is zero, therefore, property 4 of § 4 cannot be applied. Since the numerator is a constant number, and the denominator [x2x) -> -0 as x - - 1, then the whole fraction increases in absolute value without limit, i.e. lim " 1 X - * - - 1 x * + x Example 4. Find lim \-ll * "!" "" The limit of the denominator is zero: lim (xr-6lg + 8) \u003d 2 * -6-2 + 8 \u003d 0, therefore X property 4 § 4 not applicable. But the limit of the numerator is also equal to zero: lim (х2 - 5d; + 6) = 22 - 5-2-f 6 = 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 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" therefore, xr- -f- 6 r x-3 -1 1 Example 5. Find lim xn (n is an integer, positive). X with We have xn \u003d X * X. . X, n times Since each factor grows indefinitely, the product also grows indefinitely, i.e., lim xn = oo. x oo Example 6. Find lim xn(n is an integer, positive). X -> - CO We have xn = x x... x. 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 ti uTL xm I lim lim lim m. X - O x-* u L X -\u003e w Here the numerator remains constant, and the denominator grows in absolute value, therefore lim -b = 0. X-*oo X* The result of this example is recommended to be remembered in the following form: The power function grows the faster, the greater the exponent. $хв_Зхг + 7

Examples

Example 8. Find lim g L -r- \u003d. In this example, x-*® "J * "G bX -ox-o and the numerator and denominator increase indefinitely. Divide both the numerator and denominator by the highest power of x, i.e. on xv, then 3 7_ Example 9. Find lira... Making transformations, we get lira... ^ = lim X CO + 3 7 3 Since lim -5 = 0, lim -, = 0, then the limit of the denominator rade-*® X X-+-CD X veins zero, while the limit of the numerator is 1. Therefore, the whole fraction increases without limit, i.e. t.7x hm X-+ u Example 10. Find lim 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 §

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 accumulated 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.

Decision. 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

Decision. 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. . To find .

Decision.

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. To find ( ).

Decision. 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.

Decision. 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.

Decision. 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.

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.

By the way! For our readers there is now a 10% discount on any kind of work

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.