1 and 2 wonderful. First and second wonderful limit
Term " wonderful limit"is widely used in textbooks and methodological manuals To designate important identities that help substantially simplify work Upon finding limits.
But to clear lead Its limit to the remarkable, you need to look close to him, because they are not in direct form, but often in the form of consequences, equipped with additional terms and multipliers. However, first theory, then examples, and everything will succeed!
First wonderful limit
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The first wonderful limit is written so (uncertainty of the type $ 0/0 $):
$$ \\ LIM \\ Limits_ (X \\ To 0) \\ FRAC (\\ Sin x) (x) \u003d 1. $$.
Consequences from the first wonderful limit
$$ \\ LIM \\ Limits_ (X \\ To 0) \\ FRAC (X) (\\ Sin x) \u003d 1. $$$$ \\ LIM \\ Limits_ (x \\ to 0) \\ FRAC (\\ Sin (AX)) (\\ sin (BX)) \u003d \\ FRAC (A) (B). $$$$ \\ LIM \\ Limits_ (x \\ to 0) \\ FRAC (\\ Tan x) (x) \u003d 1. $$$$ \\ LIM \\ Limits_ (x \\ to 0) \\ FRAC (\\ arcsin x) (x) \u003d 1. $$$$ \\ LIM \\ Limits_ (X \\ to 0) \\ FRAC (\\ arctan x) (x) \u003d 1. $$$$ \\ LIM \\ Limits_ (X \\ To 0) \\ FRAC (1- \\ COS X) (x ^ 2/2) \u003d 1. $$.Examples of solutions: 1 Wonderful limit
Example 1. Calculate the $$ \\ LIM \\ Limits_ limit (x \\ to 0) \\ FRAC (\\ SIN 3X) (8x). $$
Decision. The first step is always the same - we substitute the limit value of $ x \u003d 0 $ to the function and get:
$$ \\ LEFT [\\ FRAC (\\ Sin 0) (0) \\ Right] \u003d \\ left [\\ FRAC (0) (0) \\ RIGHT]. $$
Received the uncertainty of the type $ \\ left [\\ FRAC (0) (0) \\ Right] $, which should be disclosed. If you look attentively, the initial limit is very similar to the first wonderful, but does not coincide with it. Our task is to bring to the similarity. We convert so - we look at the expression under sine, we do the same in the denominator (conventionally speaking, multiplied and divided by $ 3x $), then cut and simplify:
$$ \\ LIM \\ Limits_ (X \\ To 0) \\ FRAC (\\ Sin 3X) (8x) \u003d \\ Limits_ (X \\ to 0) \\ FRAC (\\ Sin 3x) (3x) \\ FRAC (3x) (8x ) \u003d \\ LIM \\ Limits_ (x \\ to 0) \\ FRAC (\\ sin (3x)) (3x) \\ FRAC (3) (8) \u003d \\ FRAC (3) (8). $$.
Above the first wonderful limit: $$ \\ LIM \\ Limits_ (x \\ to 0) \\ FRAC (\\ sin (3x)) (3x) \u003d \\ LIM \\ Limits_ (Y \\ to 0) \\ FRAC (\\ sin ( y)) (y) \u003d 1, \\ Text (made a conditional replacement) y \u003d 3x. $$. Answer: $3/8$.
Example 2. Calculate the $$ \\ LIM \\ Limits_ limit (X \\ To 0) \\ FRAC (1- \\ COS 3X) (\\ Tan 2x \\ CDOT \\ SIN 4X). $$
Decision. We substitute the limit value of $ x \u003d 0 $ to the function and get:
$$ \\ LEFT [\\ FRAC (1- \\ COS 0) (\\ Tan 0 \\ CDOT \\ SIN 0) \\ Right] \u003d \\ left [\\ FRAC (1-1) (0 \\ Cdot 0) \\ Right] \u003d \\ Left [\\ FRAC (0) (0) \\ RIGHT]. $$
Received the uncertainty of the type $ \\ left [\\ FRAC (0) (0) \\ Right] $. We transform the limit using the first wonderful limit in simplification (three times!):
$$ \\ LIM \\ Limits_ (X \\ To 0) \\ FRAC (1- \\ COS 3X) (\\ Tan 2X \\ CDOT \\ SIN 4X) \u003d \\ LIM \\ Limits_ (X \\ To 0) \\ FRAC (2 \\ Sin ^ 2 (3x / 2)) (\\ sin 2x \\ Cdot \\ Sin 4x) \\ CDOT \\ COS 2x \u003d $$$$ \u003d 2 \\ LIM \\ Limits_ (x \\ to 0) \\ FRAC (\\ Sin ^ 2 (3x / 2) ) ((3x / 2) ^ 2) \\ CDOT \\ FRAC (2x) (\\ sin 2x) \\ CDOT \\ FRAC (4X) (\\ Sin 4x) \\ CDOT \\ FRAC ((3x / 2) ^ 2) (2x \\ ) \\ CDOT \\ COS 2X \u003d 2 \\ CDOT \\ FRAC (9) (32) \\ LIM \\ Limits_ (x \\ to 0) \\ cos 2x \u003d \\ FRAC (9) (16). $$.
Answer: $9/16$.
Example 3. Find a limit $$ \\ LIM \\ Limits_ (X \\ To 0) \\ FRAC (\\ sin (2x ^ 3 + 3x)) (5x-x ^ 5). $$
Decision. And what if under trigonometric function is a complex expression? Not trouble, and here we act likewise. First, check the type of uncertainty, we substitute $ x \u003d 0 $ to the function and get:
$$ \\ LEFT [\\ FRAC (\\ sin (0 + 0)) (0-0) \\ Right] \u003d \\ left [\\ FRAC (0) (0) \\ RIGHT]. $$
Received the uncertainty of the type $ \\ left [\\ FRAC (0) (0) \\ Right] $. Multiply and divide $ 2x ^ 3 + 3x $:
$$ \\ LIM \\ Limits_ (x \\ to 0) \\ FRAC (\\ sin (2x ^ 3 + 3x)) (5x-x ^ 5) \u003d \\ Limits_ (x \\ to 0) \\ FRAC (\\ Sin (2x ^ 3 + 3X)) ((2x ^ 3 + 3x)) \\ CDOT \\ FRAC (2x ^ 3 + 3X) (5x-x ^ 5) \u003d \\ Lim \\ Limits_ (x \\ to 0) 1 \\ CDOT \\ FRAC ( 2x ^ 3 + 3x) (5x-x ^ 5) \u003d \\ left [\\ FRAC (0) (0) \\ Right] \u003d $$
Again received uncertainty, but in this case it's just a fraction. Sperate by $ x $ Number and denominator:
$$ \u003d \\ LIM \\ Limits_ (x \\ to 0) \\ FRAC (2x ^ 2 + 3) (5-x ^ 4) \u003d \\ left [\\ FRAC (0 + 3) (5-0) \\ Right] \u003d \\ $$.
Answer: $3/5$.
The second wonderful limit
The second wonderful limit is written so (uncertainty of the type of $ 1 ^ \\ infty $):
$$ \\ LIM \\ Limits_ (x \\ to \\ infty) \\ left (1+ \\ FRAC (1) (X) \\ RIGHT) ^ (x) \u003d E, \\ Quad \\ Text (or) \\ QUAD \\ LIM \\ LIMITS_ ( x \\ to 0) \\ left (1 + x \\ right) ^ (1 / x) \u003d e. $$.
Consequences of the second remarkable limit
$$ \\ LIM \\ LIMITS_ (X \\ To \\ INFTY) \\ left (1+ \\ FRAC (A) (X) \\ RIGHT) ^ (BX) \u003d E ^ (AB). $$$$ \\ LIM \\ Limits_ (x \\ to 0) \\ FRAC (\\ ln (1 + x)) (x) \u003d 1. $$$$ \\ LIM \\ Limits_ (x \\ to 0) \\ FRAC (E ^ x -1) (x) \u003d 1. $$$$ \\ LIM \\ Limits_ (x \\ to 0) \\ FRAC (A ^ X-1) (X \\ LN a) \u003d 1, a\u003e 0, a \\ ne 1. $$$$ \\ LIM \\ Limits_ ( X \\ to 0) \\ FRAC ((1 + x) ^ (a) -1) (AX) \u003d 1. $$.Examples of solutions: 2 Wonderful limit
Example 4. Find a limit $$ \\ LIM \\ Limits_ (X \\ To \\ infty) \\ left (1- \\ FRAC (2) (3X) \\ RIGHT) ^ (x + 3). $$
Decision. Check the type of uncertainty, we substitute $ x \u003d \\ infty $ to the function and get:
$$ \\ left [\\ left (1- \\ FRAC (2) (\\ INFTY) \\ RIGHT) ^ (\\ INFTY) \\ RIGHT] \u003d \\ left. $$
Received the uncertainty of the type of $ \\ left $. The limit can be reduced to the second wonderful one. We transform:
$$ \\ LIM \\ Limits_ (X \\ To \\ infty) \\ left (1- \\ FRAC (2) (3X) \\ RIGHT) ^ (X + 3) \u003d \\ LIM \\ Limits_ (x \\ to \\ infty) \\ left ( 1+ \\ FRAC (1) ((- 3x / 2)) \\ right) ^ (\\ FRAC (-3x / 2) (- 3x / 2) (x + 3)) \u003d $$$$ \u003d \\ LIM \\ Limits_ (x \\ to \\ infty) \\ left (\\ left (1+ \\ FRAC (1) ((- 3x / 2)) \\ Right) ^ ((- 3x / 2)) \\ Right) ^ \\ FRAC (X + 3 ) (- 3x / 2) \u003d $$
The expression in brackets is actually the second wonderful limit of $ \\ limits_ (t \\ to \\ infty) \\ left (1+ \\ FRAC (1) (T) \\ RIGHT) ^ (T) \u003d E $, only $ T \u003d - 3X / $ 2, so
$$ \u003d \\ LIM \\ LIMITS_ (X \\ To \\ INFTY) \\ left (E \\ Right) ^ \\ FRAC (X + 3) (- 3x / 2) \u003d \\ LIM \\ Limits_ (X \\ To \\ Infty) E ^ \\ $$.
Answer: $ E ^ (- 2/3) $.
Example 5. Find $$ \\ LIM \\ Limits_ (x \\ To \\ infty) \\ left (\\ FRAC (X ^ 3 + 2X ^ 2 + 1) (x ^ 3 + x-7) \\ Right) ^ (x). $$
Decision. We substitute $ x \u003d \\ infty $ into the function and get the uncertainty of the type of $ \\ left [\\ FRAC (\\ INFTY) (\\ INFTY) \\ Right] $. And we need $ \\ left $. Therefore, let's start with the transformation of expression in brackets:
$$ \\ LIM \\ Limits_ (X \\ To \\ infty) \\ left (\\ FRAC (x ^ 3 + 2x ^ 2 + 1) (x ^ 3 + x-7) \\ Right) ^ (x) \u003d \\ Lim \\ Limits_ (x \\ to \\ infty) \\ left (\\ FRAC (X ^ 3 + (X-7) - (X-7) + 2x ^ 2 + 1) (x ^ 3 + x-7) \\ Right) ^ (x ) \u003d \\ l \\ limits_ (x \\ to \\ infty) \\ left (\\ FRAC ((x ^ 3 + x-7) + (- x + 7 + 2x ^ 2 + 1)) (x ^ 3 + x-7 ) \\ Right) ^ (x) \u003d $$$$ \u003d \\ Lim \\ limits_ (x \\ to \\ infty) \\ left (1+ \\ FRAC (2x ^ 2-x + 8) (x ^ 3 + x-7) \\ Right) ^ (x) \u003d \\ Lim \\ limits_ (x \\ to \\ infty) \\ left (\\ left (1+ \\ FRAC (2x ^ 2-x + 8) (x ^ 3 + x-7) \\ Right) ^ (\\ FRAC (x ^ 3 + x-7) (2x ^ 2-x + 8)) \\ right) ^ (x \\ FRAC (2x ^ 2-x + 8) (x ^ 3 + x-7)) \u003d $$.
The expression in brackets is actually the second wonderful limit of $ \\ limits_ (t \\ to \\ infty) \\ left (1+ \\ FRAC (1) (T) \\ RIGHT) ^ (T) \u003d E $, only $ T \u003d \\ $$ \u003d \\ LIM \\ Limits_ (x \\ to \\ infty) \\ left (E \\ Right) ^ (X \\ FRAC (2x ^ 2-x + 8) (x ^ 3 + x-7)) \u003d \\ Lim \\ Limits_ (x \\ to \\ infty) E ^ (\\ FRAC (2x ^ 2-x + 8) (x ^ 2 + 1-7 / x)) \u003d \\ Limits_ (x \\ to \\ infty) E ^ (\\ FRAC (2-1 / x + 8 / x ^ 2) (1 + 1 / x ^ 2-7 / x ^ 3)) \u003d E ^ (2). $$.
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.
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!
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 a significant-power function, but 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) $ |
| 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: |
| Decision |
|
$$ \\ LIM_ (X \\ TO \\ INFTY) \\ Bigg (1+ \\ FRAC (3X ^ 2 + 4) (3x ^ 2-2) -1 \\ Bigg) ^ (3x) \u003d \\ Lim_ (X \\ To \\ Infty ) \\ Bigg (1+ \\ FRAC (6) (3x ^ 2-2) \\ Bigg) ^ (3x) \u003d $$ 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.
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
