Wonderful limits and consequences of them. First wonderful limit

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 we 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 for.

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.
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Find wonderful limits It is difficult not only to many students of the first, second courses of study that study the theory of limits, but also to some teachers.

Formula of the first remarkable limit

Consequences of the first wonderful limit we write formulas
1. 2. 3. 4. But by themselves the general formulas of the wonderful limits of anyone on the exam or test do not help. The bottom line is that real tasks are built so that the formulas recorded above must still come. And most students who miss the pairs, study this course in absentia or have teachers who themselves do not always understand what they explain about, cannot calculate the most elementary examples for wonderful limits. From the formulas of the first wonderful limit, we see that with their help you can explore the uncertainty of the type zero divide to zero for expressions with trigonometric functions. Consider first a number of examples on the first wonderful limit, and then we study the second wonderful limit.

Example 1. Find the SIN function limit (7 * x) / (5 * x)
Solution: As you can see the function under the limit close to the first wonderful limit, but the limit of the function is not exactly equal to one. In such assignments, it follows the limits in the denominator to highlight the variable with the same coefficient, which is contained with a variable under sine. In this case, it should be divided and multiplied by 7

Some such detail will seem unnecessary, but most students who are difficult to give limits to help better understand the rules and assimilate theoretical Material.
Also, if there is a reverse type of function - it is also the first wonderful limit. And all because the wonderful limit is equal to one

The same rule concerns the consequences of 1 remarkable limit. Therefore, if you are asked "What is the first wonderful limit?" You should not answer that this is a unit.

Example 2. Find the SIN function limit (6x) / Tan (11x)
Solution: To understand the final result, with a function in the form

To apply the rules of the wonderful limit to multiply and divide the factors

Next, the limit of the function of the functions by scattering through the product limits

Without complex formulas, we found the limit of Chasska trigonometric functions. For assimilation simple formulas Try to come up and find the limit on 2 and 4 formula of the investigation 1 of the wonderful limit. We will consider more complex tasks.

Example 3. Calculate the limit (1-COS (X)) / x ^ 2
Solution: When checking the substitution, we obtain the uncertainty 0/0. Many are unknown how to reduce such an example to 1 wonderful limit. Here should be used trigonometric formula

At the same time, the limit will be transformed to understandable

We managed to reduce the function to the square of the wonderful limit.

Example 4. Find the limit
Solution: When substitution, we get a familiar feature 0/0. However, the variable tends to PI, and not to zero. Therefore, to use the first remarkable limit, we will perform such a replacement of the variable x so that the new variable is sent to zero. For this, the denominator is denoted by a new variable pi-x \u003d y

Thus, using the trigonometric formula, which is shown in the previous task, the example is reduced to 1 remote limit.

Example 5. Calculate the limit
Solution: It is first unclear how to simplify the limits. But once there is an example, then there must be an answer. The fact that the variable is sent to one gives when substitution feature of the type of zero multiply to infinity, so the tangent needs to be replaced by the formula

After that, we obtain the desired uncertainty 0/0. Next, we perform the replacement of variables in the limit, and use the frequency of Kotangens

Recent replacements allow us to use a consequence of 1 remarkable limit.

The second wonderful limit is equal to exponential

This is a classic to which in real tasks to limits is not always easy to come.
In the calculations you will need limits - the investigations of the second remarkable limit:
1. 2. 3. 4.
Due to the second remarkable limit and its consequences, it is possible to explore the uncertainty of the type of zero divided into zero, the unit to the extent infinity, and the infinity is divided into infinity, and even in the same extent

Let's start to get acquainted with simple examples.

Example 6. Find a function limit
Solution: Directly apply 2 Wonderful limit will not work. First, the indicator should be turned into a view to the term in brackets.

This is the technique of information to the 2 wonderful limit and in essence 2 formulas of the consequence of the limit.

Example 7. Find a function limit
Solution: We have tasks for 3 formula of the investigation 2 of the remarkable limit. Zero substitution gives a feature of 0/0. To build the limit to the rule, turn the denominator, so that with the variable there was the same coefficient as in the logarithm

It is also easy to understand and execute on the exam. Student difficulties when calculating limits begin with the following tasks.

Example 8. Calculate the limit of the function [(x + 7) / (x-3)] ^ (x-2)
Solution: We have a feature of type 1 to the degree of infinity. If you do not believe, you can substitute infinity instead of "X" and make sure that. To build a rule, we divide the numerator to the denominator in brackets, for this pre-perform manipulation

Substitute the expression on the limit and turn to 2 wonderful limit

The limit is equal to the exhibitor at 10 degrees. Constants, which are the terms with a variable as in brackets and do not make any "weather" - should be remembered about it. And if you are asked teachers - "why not turn the indicator?" (For this example in X-3), then say that "when the variable tends to infinity, then add 100 at least 1000 to it, and the limit will remain as it was!".
There is a second way to calculate the limits of this type. Tell about him in the next task.

Example 9. Find a limit
Solution: Now we will bring a variable in the numerator and the denominator and turn the ORD feature to another. To obtain the final value, we use the Formula of the investigation 2 of the remarkable limit

Example 10. Find a function limit
Solution: the specified limit is not found at all. For the construction of under 2 limit, we will imagine that Sin (3x) is a variable, and you need to turn the indicator

Next, we write as a degree to degree


In brackets described intermediate reasoning. As a result of the use of the first and second remarkable limit, they received the exhibitor in Cuba.

Example 11. Calculate the limit of the function sin (2 * x) / ln (3 * x + 1)
Solution: We have uncertainty of the type 0/0. In addition, we see that the function should be turned to the use of both wonderful limits. Perform previous mathematical transformations

Next easily the limit will take a value

This is how freely you will feel in tests, tests, modules if you learn how to quickly write functions and reduce the first or second wonderful limit. If you learn the above methods of finding limits is difficult for you, you can always order test At our limits.
To do this, fill out the form, specify the data and attach the file with examples. We helped many students - we can help you!

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, right part will increase, 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. Each value of X is concluded between two positive integers: where is whole 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 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

This article: "The second wonderful limit" is devoted to disclosure within the limits of uncertainty of the form:

$ \\ Bigg [\\ FRAC (\\ INFTY) (\\ INFTY) \\ Bigg] ^ \\ infty $ and $ ^ \\ infty $.

Also, such uncertainties can be disclosed by logarithming power functionBut this is another solution method that will be covered in another article.

Formula and investigation

Formula The second remarkable limit is written as follows: $$ \\ LIM_ (X \\ To \\ infty) \\ Bigg (1+ \\ FRAC (1) (X) \\ Bigg) ^ x \u003d E, \\ Text (where) E \\ APPROX 2.718 $$

From the formula flow out corollarywhich are very convenient to apply to solve examples from the limits: $$ \\ Lim_ (x \\ to \\ infty) \\ Bigg (1 + \\ FRAC (K) (X) \\ Bigg) ^ x \u003d E ^ k, \\ Text (where) k \\ in \\ mathbb (r) $$$$ \\ lim_ (x \\ to \\ infty) \\ Bigg (1 + \\ FRAC (1) (F (X)) \\ Bigg) ^ (f (x)) \u003d E $ $ $$ \\ LIM_ (X \\ To 0) \\ Bigg (1 + x \\ Bigg) ^ \\ FRAC (1) (X) \u003d E $$

It costs to notice that the second wonderful limit can not be applied to an indicative-power function, but only in cases when the base is committed to one. To do this, first in the mind calculate the base limit, and then draw conclusions. All this will be considered in the examples of solutions.

Examples of solutions

Consider examples of solutions using the direct formula and its consequences. We also analyze cases in which the formula is not needed. Just write only the finished answer.

Example 1.

Find the limit of $ \\ lim_ (x \\ to \\ infty) \\ Bigg (\\ FRAC (X + 4) (X + 3) \\ Bigg) ^ (x + 3) $

Decision

Substitute infinity to the limit and look at uncertainty: $$ \\ Lim_ (X \\ To \\ infty) \\ Bigg (\\ FRAC (X + 4) (X + 3) \\ Bigg) ^ (x + 3) \u003d \\ Bigg (\\ FRAC (\\ INFTY) (\\ INFTY) \\ Bigg) ^ \\ INFTY $$ Find the base limit: $$ \\ lim_ (x \\ to \\ infty) \\ FRAC (X + 4) (X + 3) \u003d \\ LIM_ (X \\ To \\ INFTY) \\ FRAC (X (1+ \\ FRAC (4) ( x))) (x (1+ \\ FRAC (3) (x))) \u003d 1 $$ Received a reason equal to one, and this means you can already apply the second wonderful limit. To do this, roag the base of the function under the formula by subtracting and adding a unit: $$ \\ LIM_ (X \\ To \\ Infty) \\ Bigg (1 + \\ FRAC (X + 4) (X + 3) - 1 \\ Bigg) ^ (x + 3) \u003d \\ Lim_ (X \\ To \\ infty) \\ We look at the second consequence and write the answer: $$ \\ LIM_ (X \\ TO \\ INFTY) \\ Bigg (1 + \\ FRAC (1) (X + 3) \\ Bigg) ^ (x + 3) \u003d E $$ If it is impossible to solve your task, then send it to us. We will provide Detailed solution . You can familiarize yourself with the course of calculation and learn information. This will help in a timely manner at the teacher!Answer

$$ \\ LIM_ (X \\ TO \\ INFTY) \\ Bigg (1 + \\ FRAC (1) (X + 3) \\ Bigg) ^ (x + 3) \u003d E $$

Example 4.

Solve the limit of $ \\ lim_ (x \\ to \\ infty) \\ Bigg (\\ FRAC (3x ^ 2 + 4) (3x ^ 2-2) \\ Bigg) ^ (3x) $

Decision

We find the base limit and see that $ \\ lim_ (x \\ to \\ infty) \\ FRAC (3x ^ 2 + 4) (3x ^ 2-2) \u003d 1 $, then you can apply the second wonderful limit. The standard according to the plan add and subtract a unit from the foundation of the degree: $$ \\ LIM_ (X \\ TO \\ INFTY) \\ Bigg (1+ \\ FRAC (3X ^ 2 + 4) (3x ^ 2-2) -1 \\ Bigg) ^ (3x) \u003d \\ Lim_ (X \\ To \\ Infty ) \\ Bigg (1+ \\ FRAC (6) (3x ^ 2-2) \\ Bigg) ^ (3x) \u003d $$ We climb the fraction under the formula of the 2nd note. limit: $$ \u003d \\ LIM_ (X \\ To \\ INFTY) \\ Bigg (1+ \\ FRAC (1) (\\ FRAC (3X ^ 2-2) (6)) \\ Bigg) ^ (3x) \u003d $$ Now we customize the degree. A degree should be a fraction equal to the denominator of the base $ \\ FRAC (3x ^ 2-2) (6) $. For this, multiply and divide the degree on it, and continue to decide: $$ \u003d \\ LIM_ (X \\ TO \\ INFTY) \\ Bigg (1+ \\ FRAC (1) (\\ FRAC (3X ^ 2-2) (6)) \\ Bigg) ^ (\\ FRAC (3x ^ 2-2) (6) \\ CDOT \\ FRAC (6) (3x ^ 2-2) \\ CDOT 3X) \u003d \\ LIM_ (X \\ To \\ INFTY) E ^ (\\ FRAC (18x) (3x ^ 2-2)) \u003d $$ The limit located to the degree at $ E $ is: $ \\ Lim_ (x \\ to \\ infty) \\ FRAC (18x) (3x ^ 2-2) \u003d 0 $. Therefore, continuing the decision:

$$ \\ LIM_ (X \\ TO \\ INFTY) \\ Bigg (1 + \\ FRAC (1) (X + 3) \\ Bigg) ^ (x + 3) \u003d E $$

$$ \\ LIM_ (X \\ To \\ INFTY) \\ Bigg (\\ FRAC (3X ^ 2 + 4) (3x ^ 2-2) \\ Bigg) ^ (3x) \u003d 1 $$

We analyze cases when the task is similar to the second wonderful limit, but is solved without it.

In the article: "The second wonderful limit: examples of solutions" was dismantled by the formula, its consequence and the frequent types of tasks are given on this topic.

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.

Yandex.rtb R-A-339285-1

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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