Evidence:

We first prove the theorem for the case of a sequence

According to the Binoma Newton formula:

Personal reception

From this equality (1) it follows that with increasing N, the number of positive terms in the right part increases. In addition, with an increase in n, the number decreases, so the values increase. Therefore, the sequence increasing, with (2) * We show that it is limited. I will replace each bracket in the right part of equality per unit, the right side will increase, we get inequality

We will replace the resulting inequality, replace 3,4,5, ... who are in denominators of fractions, number 2: the amount in the bracket will find by the formula of the sum of the members of the geometric progression: therefore (3)*

So, the sequence is limited from above, at the same time inequalities (2) and (3) are performed: Consequently, on the basis of Weierstrass theorem (sequence convergence criterion) sequence Monotonously increases and limited, it means that the limit is indicated by the letter E. Those.

Knowing that the second wonderful limit is faithful for natural values \u200b\u200bx, we will prove the second wonderful limit for real X, that is, we prove that . Consider two cases:

1. Let each value of X are concluded between two positive integers: where is the integer part x. \u003d\u003e \u003d\u003e

If, therefore, according to the limit Have

In the sign (about the limit of the intermediate function) of the existence of the limits

2. Let. Make a substitution - x \u003d t, then

Of these two cases, it follows that For real X.

Corollary:

9 .) Comparison is infinitely small. The replacement theorem is infinitely small to equivalent in the limit and the theorem on the main part of infinitely small.

Let functions a ( x.) and b ( x.) - B.M. for x. ® x. 0 .

Definitions.

1) A ( x.) called infinitely low higher order than b. (x.) if a

Record: A ( x.) \u003d O (B ( x.)) .

2) A ( x.) andb ( x.) called infinitely small one order, if a

where S.ℝℝ I. C.¹ 0 .

Record: A ( x.) = O.(B ( x.)) .

3) A ( x.) and b ( x.) called equivalent , if a

Record: A ( x.) ~ B ( x.).

4) A ( x.) called infinitely small order k
Infinitely smallb ( x.), If infinitely smalla ( x.) and(B ( x.)) K. have one order, i.e. if a

where S.ℝℝ I. C.¹ 0 .

THEOREM 6 (On the replacement of infinitely small on the equivalent).

Let bea ( x.), b ( x.), a 1 ( x.), b 1 ( x.) - B.M. With X. ® x. 0 . If aa ( x.) ~ A 1 ( x.), b ( x.) ~ B 1 ( x.),

that

Proof: Let A ( x.) ~ A 1 ( x.), b ( x.) ~ B 1 ( x.), then

THEOREM7 (about the main part of infinitely small).

Let bea ( x.) andb ( x.) - B.M. With X. ® x. 0 , andb ( x.) - B.M. Higher order thana ( x.).

\u003d, A since b ( x.) - Higher order than A ( x.), i.e. of It is clear that A ( x.) + b ( x.) ~ A ( x.)

10) The continuity of the function at the point (in the language of the epsilon-delta limits, geometric) one-sided continuity. Continuity on the interval, on the segment. Properties of continuous functions.

1. Basic definitions

Let be f.(x.) defined in some neighborhood of the point x. 0 .

Definition 1. Function F.(x.) called continuous at point x. 0 if equality is right

Remarks.

1) by virtue of Theorem 5 §3 Equality (1) can be written as

Condition (2) - determining the continuity of the function at the point in the language of one-way limits.

2) Equality (1) can also be written as:

They say: "If the function is continuous at the point x. 0, then the limit sign and the function can be changed in places. "

Definition 2 (in E-D).

Function F.(x.) called continuous at point x. 0 if a "E\u003e 0 $ D\u003e 0 that, what

If X.Îu ( x. 0, D) (i.e. | x. – x. 0 | < d),

that F.(x.) Îu ( f.(x. 0), E) (i.e. | f.(x.) – f.(x. 0) | < e).

Let be x., x. 0 Î D.(f.) (x. 0 - fixed, x -arbitrary)

Denote: D. x.= x - X. 0 – argument increment

D. f.(x. 0) = f.(x.) – f.(x. 0) – protect function at pointX 0

Definition 3 (geometric).

Function F.(x.) on the turns out continuous at point x. 0 if at this point the infinitely small increment of the argument corresponds to the infinitely small increment of the function.

Let the function f.(x.) determined on the interval [ x. 0 ; x. 0 + d) (on the interval ( x. 0 - D; x. 0 ]).

Definition. Function F.(x.) called continuous at point x. 0 on right (left ), if equality is right

It's obvious that f.(x.) continuous at point x. 0 Û f.(x.) continuous at point x. 0 Right and left.

Definition. Function F.(x.) called continuous on interval e ( a.; b.) if it is continuous at every point of this interval.

Function F.(x.) called continuous on the segment [a.; b.] if it is continuous on the interval (a.; b.) and has one-sided continuity at boundary points (i.e. continuous at the point a. right at point b. - Left).

11) Rippoints, their classification

Definition. If F.(x.) defined in some neighborhood point x 0 , but it is not continuous at this point, f.(x.) call the discontinuous point x 0 , and the point itself x. 0 call a point of break Functions F.(x.) .

Remarks.

1) f.(x.) can be determined in an incomplete neighborhood of the point x. 0 .

Then consider the corresponding one-sided continuity of the function.

2) from the definition þ point x. 0 is a function break point f.(x.) In two cases:

a) U ( x. 0, d) î D.(f.) , but for f.(x.) Equality is not performed

b) u * ( x. 0, d) î D.(f.) .

For elementary functions, only a case b is possible.

Let be x. 0 - function break point f.(x.) .

Definition. Point X. 0 called spray point I. roda If F.(x.) has at this point the end limits on the left and right.

If this limits are equal, then point x 0 called disposable break point , otherwise - point of jump .

Definition. Point X. 0 called spray point II. roda If at least one of the one-sided limits of the function f(x.) At this point is equal¥ or does not exist.

12) Properties of functions continuous on the segment (Weierstrass theorems (without docking) and Cauchy

Weierstrass theorem

Let the function f (x) be continuous on the segment, then

1) F (x) is limited to

2) F (X) takes its smallest and most important

Definition: The value of the function m \u003d fits the smallest if M≤f (x) for any X € D (F).

The value of the function m \u003d fits the greatest if M≥f (x) for any X € D (F).

The smallest \\ the greatest value can take at several segments.

f (x 3) \u003d f (x 4) \u003d max

Cauchy theorem.

Suppose that the function f (x) is continuous on the segment and x - the number concluded between F (a) and F (b), then there is at least one point x 0 € such that f (x 0) \u003d G

From the above article you can find out what is the limit, and what it is eaten - it is very important. Why? You can not understand what determinants and successfully solve them, you can absolutely not understand what is derived and find them on the "five". But if you do not understand what the limit is, the decision of the practical tasks will have to be tight. It will not be superfluous to familiarize themselves with the samples of decisions and my design recommendations. All information is set forth in a simple and accessible form.

And for the purposes of this lesson, we will need the following methodological materials: Wonderful limits and Trigonometric formulas. They can be found on the page. It is best to print the methods - it is much more convenient, moreover, they will often have to contact them.

What are wonderful remarkable limits? The remarks of these limits are that they are proven by the greatest minds of famous mathematicians, and the grateful descendants do not have to suffer terrible limits with the journey of trigonometric functions, logarithms, degrees. That is, when you find the limits, we will use the finished results, which are proven theoretically.

There are several wonderful limits, but in practice, in 95% of cases, two wonderful limits appear in 95% of cases: First wonderful limit, The second wonderful limit. It should be noted that these are the historically established names, and when, for example, they say about the "first remarkable limit," then they imply a completely definite thing, and not some random, taken from the ceiling limit.

First wonderful limit

Consider the next limit: (instead of the native letter "He" I will use greek letter "Alpha" is more convenient from the point of view of the material supply).

According to our rule of location (see Article Limits. Examples of solutions) We try to substitute zero to function: in the numerator, we turn out zero (zero sine equal to zero.), in the denominator, obviously also zero. Thus, we are faced with the uncertainty of the species, which, fortunately, is not necessary to disclose. In the course of mathematical analysis, it is proved that:

This mathematical fact is called The first wonderful limit. Analytical proof of the limit will not bring, but its geometrical meaning will look at the lesson about infinitely small features.

Often in practical tasks, the functions can be located differently, it does not change anything:

- The same first wonderful limit.

But independently rearrange the numerator and the denominator can not! If there is a limit in the form, then it is necessary to solve it in the same form, without rearring.

In practice, not only a variable, but also an elementary function, complex function can act as a parameter. It is important only to silent to zero.

Examples:
, , ,

Here , , , , And all Hood is the first wonderful limit apply.

But the next post is heresy:

Why? Because the polynomial does not seek zero, he strives for the top five.

By the way, the question of snowing, and what is equal to the limit ? The answer can be found at the end of the lesson.

In practice, not everything is so smooth, almost never a student will be offered to solve the freezing limit and get a light offset. Hmmm ... I write these lines, and a very important thought occurred - all the same "free" mathematical definitions and formulas seems to be better remembered by heart, it can have invaluable assistance on the competition, when the issue will be solved between the "Two" and "Troika" and teacher will decide to ask a student any simple question or suggest solving the simplest example ("Maybe he (a) still knows what?!").

We proceed to consideration of practical examples:

Example 1.

Find a limit

If we notice in the limit of Sinus, then this should immediately come out for the idea of \u200b\u200busing the first remarkable limit.

First try to substitute 0 in the expression under the sign of the limit (we do it mentally or on the draft):

So, we have uncertainty of the species, her we definitely indicate In deciding the solution. The expression under the sign of the limit seems to be the first wonderful limit, but it is not at all, under sine is located, and in the denominator.

In such cases, the first wonderful limit we need to organize yourself using artificial reception. The course of reasoning can be like this: "Under sine we, it means, we also need to get in the denominator."
And this is done very simple:

That is, the denominator is artificially multiplied in this case by 7 and is divided into the same seven. Now the record has taken familiar outlines.
When the task is drawn up by hand, then the first wonderful limit is desirable to mark with a simple pencil:

What happened? In essence, we turned into a unit circled and disappeared in the work:

Now it is only left to get rid of three-storey fractions:

Who forgot the simplification of multi-storey fractions, please refresh the material in the directory Hot Mathematics School Course Formulas .

Ready. Final answer:

If you do not want to use the mark with a pencil, then the decision can be issued as follows:

“

We use the first wonderful limit.

“

Example 2.

Find a limit

Again we see in the limit fraction and sinus. We try to substitute nizer and denominator zero:

Indeed, we have uncertainty and, it means you need to try to organize the first wonderful limit. At the lesson Limits. Examples of solutions We considered the rule that when we have uncertainty, you need to decompose the numerator and denominator for multipliers. Here - the same, degrees we will present in the form of a work (multipliers):

Similar to the previous example, wonderful limits with a pencil (here are two), and we indicate that they strive for a unit:

Actually, the answer is ready:

In the following examples, I will not engage in arts in Painte, I think how to make a solution to the notebook - you already understand.

Example 3.

Find a limit

We substitute zero to the expression under the sign of the limit:

The uncertainty is obtained that you need to disclose. If there is a tangent in the limit, it is almost always converted to sinus and cosine at a well-known trigonometric formula (by the way, with a catangent, see the same thing, see Methodical Material Hot trigonometric formulas On the page Mathematical formulas, tables and reference materials).

In this case:

Cosine zero equal to unityAnd it's easy to get rid of it (do not forget to mark that he seeks a unit):

Thus, if the limit of the cosine is a multiplier, then it is rude, it is necessary to turn into a unit that disappears in the work.

Here everything came out easier, without any narrations and divisions. The first wonderful limit also turns into a unit and disappears in the work:

As a result, infinity was obtained, it happens.

Example 4.

Find a limit

We try to substitute zero into a numerator and denominator:

Uncertainty (cosine zero, as we remember, is equal to one)

Use the trigonometric formula. Take note! For some reason, the limits with the use of this formula are very often found.

Permanent multipliers will bring the limit for the icon:

We organize the first wonderful limit:

Here we have only one wonderful limit that turns into a unit and disappears in the work:

Get rid of three-storey:

The limit is actually resolved, we indicate that the remaining sine seeks zero:

Example 5.

Find a limit

This example is more difficult, try to figure it out on your own:

Some limits can be reduced to the 1st remote limit by replacing the variable, you can read a little later in the article. Methods for solving limits.

The second wonderful limit

In the theory of mathematical analysis, it is proved that:

This fact is called title the second remarkable limit.

Reference: - This is an irrational number.

As a parameter, not only a variable, but also a complex function. It is important only to strive to infinity.

Example 6.

Find a limit

When an expression under the sign of a limit is to a degree - this is the first sign that you need to try to apply the second wonderful limit.

But first, as always, we try to substitute an infinitely large number in the expression, according to which principle it is done, disassembled at the lesson Limits. Examples of solutions.

It is easy to see that when The foundation of the degree, and the indicator - That is, there is an uncertainty of the form:

This uncertainty is just revealed using the second remarkable limit. But, as often happens, the second wonderful limit does not lie on a saucer with a blue burrow, and it needs to be artificially organized. You can argue as follows: In this example, the parameter, it means, in the indicator, we also need to be organized. To do this, we will be erected into a degree, and so that the expression has not changed - we will be taken to the degree:

When the task is drawn up by hand, tagged with a pencil:

Almost everything is ready, a terrible degree has become a pretty letter:

At the same time, the limit icon itself moves to the indicator:

Example 7.

Find a limit

Attention! The limit of this type is very often found, please read this example very carefully.

We try to substitute an infinitely large number in the expression, standing under the sign of the limit:

As a result, uncertainty was obtained. But the second wonderful limit applies to the uncertainty of the species. What to do? It is necessary to convert the degree. We argue like this: in the denominator, we, it means, in the numerator, too, should be organized.

There are several wonderful limits, but the first and second wonderful limits are the most famous. The remarks of these limits are that they have widespread use and with their help you can find other limits found in numerous tasks. This will be done in the practical part of this lesson. To solve problems, by bringing to the first or second remote limit, you do not need to disclose uncertainty contained in them, since the values \u200b\u200bof these limits have long brought great mathematicians.

The first wonderful limit It is called the limit of the sine ratio of infinitely small arc to the same arc expressed in the radian extent:

Go to solving problems for the first wonderful limit. Note: if the limit is a trigonometric function, this is almost a sure sign that this expression can be brought to the first wonderful limit.

Example 1.Find a limit.

Decision. Substitution instead x. Scratch leads to uncertainty:

.

In the denominator - sinus, therefore, the expression can be brought to the first wonderful limit. We start converting:

.

In the denominator - sinus of three X, and in a numerator only one X, it means you need to get three X and in a numeric. For what? To present 3. x. = a. And get an expression.

And we come to the variety of the first wonderful limit:

because it doesn't matter what the letter (variable) in this formula is worth it instead of ix.

We multiply the X three and immediately divide:

.

In accordance with the first wonderful limit, we produce a change in fractional expression:

Now we can finally solve this limit:

.

Example 2.Find a limit.

Decision. The direct substitution again leads to the uncertainty "zero to divide to zero":

.

To get the first wonderful limit, it is necessary that the X is under the sinus sign in a numerator and simply an X denuncator with the same coefficient. Let this coefficient be equal to 2. To do this, imagine the current coefficient of ICC as further by producing action with fractions, we obtain:

.

Example 3.Find a limit.

Decision. When substitution, we again get the uncertainty "zero to divide to zero":

.

Probably, you already understand that from the initial expression you can get the first wonderful limit multiplied by the first wonderful limit. For this, we declare the squares of the ICA in a numerator and sine in the denominator to the same multipliers, and in order to receive the same coefficients from the ICS and sinus, Icuses divide by 3 and immediately multiply on 3. We get:

.

Example 4.Find a limit.

Decision. We again get the uncertainty "zero divide to zero":

.

We can get the ratio of the first first remarkable limits. We divide and numerator, and denominator on X. Then, so that the coefficients in sinuses and the focus coincides, the upper X is multiply by 2 and immediately divided by 2, and the lower X is multiplied by 3 and immediately divide on 3. We get:

Example 5.Find a limit.

Decision. And again the uncertainty "zero to divide to zero":

We remember from trigonometry that tangent is the ratio of sine to cosine, and the cosine of zero is equal to one. We produce conversion and get:

.

Example 6.Find a limit.

Decision. The trigonometric function under the sign of the limit again pursues the idea of \u200b\u200busing the first remarkable limit. We present it as a ratio of sinus to the cosine.

Now with a calm soul go to consideration wonderful limits.
It has appearance.

Instead of the variable x, various functions may be present, the main thing is that they strive to 0.

It is necessary to calculate the limit

As can be seen, this limit is very similar to the first wonderful, but it is not quite so. In general, if you notice in the SIN limit, then you need to immediately think about whether the use of the first wonderful limit is possible.

According to our rule No. 1, we substitute instead of x zero:

We get uncertainty.

Now let's try to independently organize the first wonderful limit. To do this, we will carry out a non-hard combination:

Thus, we organize a numerator and a denominator so as to highlight 7x. I have already manifested yourself with a familiar limit. It is advisable to highlight it when deciding:

Substitute the decision of the first wonderful example and get:

We simplify the fraction:

Answer: 7/3.

As you can see - everything is very simple.

Has appearance where E \u003d 2,718281828 ... is an irrational number.

Instead of the variable x, various functions may be present, the main thing is that they strive to.

It is necessary to calculate the limit

Here we see the existence under the sign of the limit, it means it is possible to use the second remarkable limit.

As always, we will use Rule No. 1 - we will substitute instead of x:

It can be seen that at least the foundation of the degree, and the indicator - 4x\u003e, i.e. We get uncertainty of the form:

We use the second wonderful limit for disclosing our uncertainty, but first it is necessary to organize it. As can be seen - it is necessary to achieve the presence in the indicator, for which they erect the base to the degree 3x, and at the same time in the degree 1 / 3x so that the expression does not change:

Do not forget to allocate our wonderful limit:

These are really wonderful limits!
If you have any questions about first and second wonderful limits, I boldly ask them in the comments.
Everyone will answer everyone.

You can also work out with the teacher on this topic.
We are pleased to offer you the selection services of a qualified tutor in your city. Our partners will quickly select a good teacher for you on favorable conditions for you.

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The formula of the second remarkable limit has the form Lim X → ∞ 1 + 1 x x \u003d e. Another form of recording looks like this: Lim X → 0 (1 + x) 1 x \u003d e.

When we talk about the second wonderful limit, we have to deal with the uncertainty of the form 1 ∞, i.e. Unit to an infinite degree.

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Consider the tasks in which we use the ability to calculate the second wonderful limit.

Example 1.

Find the limit Lim X → ∞ 1 - 2 x 2 + 1 x 2 + 1 4.

Decision

We substitute the necessary formula and perform calculations.

lim X → ∞ 1 - 2 x 2 + 1 x 2 + 1 4 \u003d 1 - 2 ∞ 2 + 1 ∞ 2 + 1 4 \u003d 1 - 0 ∞ \u003d 1 ∞

We in the answer turned out a unit to the degree of infinity. To determine the solution method, use the uncertainty table. Choose a second wonderful limit and replace variables.

t \u003d - x 2 + 1 2 ⇔ x 2 + 1 4 \u003d - T 2

If X → ∞, then T → - ∞.

Let's see what happened after replacement:

lim X → ∞ 1 - 2 x 2 + 1 x 2 + 1 4 \u003d 1 ∞ \u003d Lim X → ∞ 1 + 1 T - 1 2 T \u003d Lim T → ∞ 1 + 1 T T - 1 2 \u003d E - 1 2

Answer: Lim X → ∞ 1 - 2 x 2 + 1 x 2 + 1 4 \u003d E - 1 2.

Example 2.

Calculate the limit Lim X → ∞ x - 1 x + 1 x.

Decision

Substitute infinity and get the following.

lim X → ∞ x - 1 x + 1 x \u003d Lim X → ∞ 1 - 1 x 1 + 1 x x \u003d 1 - 0 1 + 0 ∞ \u003d 1 ∞

In response, we again turned out the same as in the previous task, therefore, we can again take advantage of the second wonderful limit. Next, we need to highlight the whole part at the base of the power function:

x - 1 x + 1 \u003d x + 1 - 2 x + 1 \u003d x + 1 x + 1 - 2 x + 1 \u003d 1 - 2 x + 1

After that, the limit acquires the following form:

lim X → ∞ x - 1 x + 1 x \u003d 1 ∞ \u003d Lim X → ∞ 1 - 2 x + 1 x

We replace variables. Suppose that T \u003d - X + 1 2 ⇒ 2 T \u003d - x - 1 ⇒ x \u003d - 2 T - 1; If X → ∞, then T → ∞.

After that, we write down that we did in the initial limit:

lim X → ∞ x - 1 x + 1 x \u003d 1 ∞ \u003d Lim X → ∞ 1 - 2 x + 1 x \u003d Lim X → ∞ 1 + 1 T - 2 T - 1 \u003d \u003d Lim X → ∞ 1 + 1 T - 2 T · 1 + 1 T - 1 \u003d Lim X → ∞ 1 + 1 T - 2 T · Lim X → ∞ 1 + 1 T - 1 \u003d \u003d Lim X → ∞ 1 + 1 TT - 2 · 1 + 1 ∞ \u003d E - 2 · (1 + 0) - 1 \u003d E - 2

To perform this transformation, we used the basic properties of limits and degrees.

Answer: Lim X → ∞ x - 1 x + 1 x \u003d e - 2.

Example 3.

Calculate the limit Lim X → ∞ x 3 + 1 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5.

Decision

lim X → ∞ x 3 + 1 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 \u003d Lim X → ∞ 1 + 1 x 3 1 + 2 x - 1 x 3 3 2 x - 5 x 4 \u003d \u003d 1 + 0 1 + 0 - 0 3 0 - 0 \u003d 1 ∞

After that, we need to convert the function to apply the second remarkable limit. We did the following:

lim X → ∞ x 3 + 1 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 \u003d 1 ∞ \u003d Lim x → ∞ x 3 - 2 x 2 - 1 - 2 x 2 + 2 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 \u003d \u003d Lim X → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5

lim X → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 \u003d Lim X → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 x 3 + 2 x 2 - 1 - 2 x 2 + 2 - 2 x 2 + 2 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 \u003d \u003d Lim X → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 x 3 + 2 x 2 - 1 - 2 x 2 + 2 - 2 x 2 + 2 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5

Since now we have the same indicators of the degree in the numerator and denomoter of the fraction (equal six), the limit of the fraction will be equal to the ratio of these coefficients in senior degrees.

lim X → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 x 3 + 2 x 2 - 1 - 2 x 2 + 2 - 2 x 2 + 2 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 \u003d \u003d Lim X → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 x 3 + 2 x 2 - 1 - 2 x 2 + 2 - 6 2 \u003d Lim X → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 x 3 + 2 x 2 - 1 - 2 x 2 + 2 - 3

When replacing T \u003d x 2 + 2 x 2 - 1 - 2 x 2 + 2, we will have a second wonderful limit. Means what:

lim X → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 x 3 + 2 x 2 - 1 - 2 x 2 + 2 - 3 \u003d Lim X → ∞ 1 + 1 TT - 3 \u003d E - 3.

Answer: Lim X → ∞ x 3 + 1 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 \u003d E - 3.

conclusions

Uncertainty 1 ∞, i.e. The unit is infinite, is a power uncertainty, therefore, it can be disclosed using the rules for finding limits of significant power functions.

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