1 and 2 are great. The first and second wonderful limit
The term " wonderful limit"is widely used in textbooks and teaching aids to denote important identities that help substantially simplify work by finding the limits.
But to be able to bring its limit to the wonderful, you need to take a good look at it, because they do not occur in direct form, but often in the form of consequences, equipped with additional terms and factors. However, first the theory, then examples, and you will succeed!
The first wonderful limit
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The first remarkable limit is written like this ($ 0/0 $ uncertainty):
$$ \ lim \ limits_ (x \ to 0) \ frac (\ sin x) (x) = 1. $$
Consequences from the first remarkable limit
$$ \ lim \ limits_ (x \ to 0) \ frac (x) (\ sin x) = 1. $$ $$ \ lim \ limits_ (x \ to 0) \ frac (\ sin (ax)) (\ sin (bx)) = \ frac (a) (b). $$ $$ \ lim \ limits_ (x \ to 0) \ frac (\ tan x) (x) = 1. $$ $$ \ lim \ limits_ (x \ to 0) \ frac (\ arcsin x) (x) = 1. $$ $$ \ lim \ limits_ (x \ to 0) \ frac (\ arctan x) (x) = 1. $$ $$ \ lim \ limits_ (x \ to 0) \ frac (1- \ cos x) (x ^ 2/2) = 1. $$Solution Examples: 1 Wonderful Limit
Example 1. Calculate limit $$ \ lim \ limits_ (x \ to 0) \ frac (\ sin 3x) (8x). $$
Solution. The first step is always the same - we substitute the limit value $ x = 0 $ into the function and we get:
$$ \ left [\ frac (\ sin 0) (0) \ right] = \ left [\ frac (0) (0) \ right]. $$
We got an uncertainty of the form $ \ left [\ frac (0) (0) \ right] $, which should be expanded. If you look closely, the original limit is very similar to the first great, but not the same. Our task is to bring to similarity. We transform this way - we look at the expression under the sine, do the same in the denominator (relatively speaking, we multiplied and divided by $ 3x $), then we reduce and simplify:
$$ \ lim \ limits_ (x \ to 0) \ frac (\ sin 3x) (8x) = \ lim \ limits_ (x \ to 0) \ frac (\ sin 3x) (3x) \ frac (3x) (8x ) = \ lim \ limits_ (x \ to 0) \ frac (\ sin (3x)) (3x) \ frac (3) (8) = \ frac (3) (8). $$
Above just got the first wonderful limit: $$ \ lim \ limits_ (x \ to 0) \ frac (\ sin (3x)) (3x) = \ lim \ limits_ (y \ to 0) \ frac (\ sin ( y)) (y) = 1, \ text (made a conditional substitution) y = 3x. $$ Answer: $3/8$.
Example 2. Calculate limit $$ \ lim \ limits_ (x \ to 0) \ frac (1- \ cos 3x) (\ tan 2x \ cdot \ sin 4x). $$
Solution. We substitute the limit value $ x = 0 $ into the function and we get:
$$ \ left [\ frac (1- \ cos 0) (\ tan 0 \ cdot \ sin 0) \ right] = \ left [\ frac (1-1) (0 \ cdot 0) \ right] = \ left [\ frac (0) (0) \ right]. $$
We got an uncertainty of the form $ \ left [\ frac (0) (0) \ right] $. Convert the limit by using the first wonderful limit (three times!) In simplification:
$$ \ lim \ limits_ (x \ to 0) \ frac (1- \ cos 3x) (\ tan 2x \ cdot \ sin 4x) = \ lim \ limits_ (x \ to 0) \ frac (2 \ sin ^ 2 (3x / 2)) (\ sin 2x \ cdot \ sin 4x) \ cdot \ cos 2x = $$ $$ = 2 \ lim \ limits_ (x \ to 0) \ frac (\ sin ^ 2 (3x / 2) ) ((3x / 2) ^ 2) \ cdot \ frac (2x) (\ sin 2x) \ cdot \ frac (4x) (\ sin 4x) \ cdot \ frac ((3x / 2) ^ 2) (2x \ cdot 4x) \ cdot \ cos 2x = $$ $$ = 2 \ lim \ limits_ (x \ to 0) 1 \ cdot 1 \ cdot 1 \ cdot \ frac ((9/4) x ^ 2) (8x ^ 2 ) \ cdot \ cos 2x = 2 \ cdot \ frac (9) (32) \ lim \ limits_ (x \ to 0) \ cos 2x = \ frac (9) (16). $$
Answer: $9/16$.
Example 3. Find the Limit $$ \ lim \ limits_ (x \ to 0) \ frac (\ sin (2x ^ 3 + 3x)) (5x-x ^ 5). $$
Solution. What if there is a complex expression under the trigonometric function? It doesn't matter, and here we act in the same way. First, we check the type of uncertainty, substitute $ x = 0 $ into the function and get:
$$ \ left [\ frac (\ sin (0 + 0)) (0-0) \ right] = \ left [\ frac (0) (0) \ right]. $$
We got an uncertainty of the form $ \ left [\ frac (0) (0) \ right] $. Multiply and divide by $ 2x ^ 3 + 3x $:
$$ \ lim \ limits_ (x \ to 0) \ frac (\ sin (2x ^ 3 + 3x)) (5x-x ^ 5) = \ lim \ limits_ (x \ to 0) \ frac (\ sin (2x ^ 3 + 3x)) ((2x ^ 3 + 3x)) \ cdot \ frac (2x ^ 3 + 3x) (5x-x ^ 5) = \ lim \ limits_ (x \ to 0) 1 \ cdot \ frac ( 2x ^ 3 + 3x) (5x-x ^ 5) = \ left [\ frac (0) (0) \ right] = $$
Again we got uncertainty, but in this case it's just a fraction. Reduce the numerator and denominator by $ x $:
$$ = \ lim \ limits_ (x \ to 0) \ frac (2x ^ 2 + 3) (5-x ^ 4) = \ left [\ frac (0 + 3) (5-0) \ right] = \ frac (3) (5). $$
Answer: $3/5$.
Second wonderful limit
The second remarkable limit is written as follows (uncertainty of the form $ 1 ^ \ infty $):
$$ \ lim \ limits_ (x \ to \ infty) \ left (1+ \ frac (1) (x) \ right) ^ (x) = e, \ quad \ text (or) \ quad \ lim \ limits_ ( x \ to 0) \ left (1 + x \ right) ^ (1 / x) = e. $$
Consequences of the second remarkable limit
$$ \ lim \ limits_ (x \ to \ infty) \ left (1+ \ frac (a) (x) \ right) ^ (bx) = e ^ (ab). $$ $$ \ lim \ limits_ (x \ to 0) \ frac (\ ln (1 + x)) (x) = 1. $$ $$ \ lim \ limits_ (x \ to 0) \ frac (e ^ x -1) (x) = 1. $$ $$ \ lim \ limits_ (x \ to 0) \ frac (a ^ x-1) (x \ ln a) = 1, a> 0, a \ ne 1. $$ $$ \ lim \ limits_ ( x \ to 0) \ frac ((1 + x) ^ (a) -1) (ax) = 1. $$Solution examples: 2 wonderful limit
Example 4. Find the limit $$ \ lim \ limits_ (x \ to \ infty) \ left (1- \ frac (2) (3x) \ right) ^ (x + 3). $$
Solution. Let's check the type of uncertainty, substitute $ x = \ infty $ into the function and get:
$$ \ left [\ left (1- \ frac (2) (\ infty) \ right) ^ (\ infty) \ right] = \ left. $$
We got an uncertainty of the form $ \ left $. The limit can be reduced to the second remarkable. Let's transform:
$$ \ lim \ limits_ (x \ to \ infty) \ left (1- \ frac (2) (3x) \ right) ^ (x + 3) = \ lim \ limits_ (x \ to \ infty) \ left ( 1+ \ frac (1) ((- 3x / 2)) \ right) ^ (\ frac (-3x / 2) (- 3x / 2) (x + 3)) = $$ = \ lim \ limits_ (x \ to \ infty) \ left (\ left (1+ \ frac (1) ((- 3x / 2)) \ right) ^ ((- 3x / 2)) \ right) ^ \ frac (x + 3 ) (- 3x / 2) = $$
The expression in brackets is actually the second wonderful limit $ \ lim \ limits_ (t \ to \ infty) \ left (1+ \ frac (1) (t) \ right) ^ (t) = e $, only $ t = - 3x / 2 $, therefore
$$ = \ lim \ limits_ (x \ to \ infty) \ left (e \ right) ^ \ frac (x + 3) (- 3x / 2) = \ lim \ limits_ (x \ to \ infty) e ^ \ frac (1 + 3 / x) (- 3/2) = e ^ (- 2/3). $$
Answer:$ e ^ (- 2/3) $.
Example 5. Find the Limit $$ \ lim \ limits_ (x \ to \ infty) \ left (\ frac (x ^ 3 + 2x ^ 2 + 1) (x ^ 3 + x-7) \ right) ^ (x). $$
Solution. We substitute $ x = \ infty $ into the function and we get an uncertainty of the form $ \ left [\ frac (\ infty) (\ infty) \ right] $. And we need $ \ left $. So let's start by converting the parenthesized expression:
$$ \ lim \ limits_ (x \ to \ infty) \ left (\ frac (x ^ 3 + 2x ^ 2 + 1) (x ^ 3 + x-7) \ right) ^ (x) = \ lim \ limits_ (x \ to \ infty) \ left (\ frac (x ^ 3 + (x-7) - (x-7) + 2x ^ 2 + 1) (x ^ 3 + x-7) \ right) ^ (x ) = \ lim \ limits_ (x \ to \ infty) \ left (\ frac ((x ^ 3 + x-7) + (- x + 7 + 2x ^ 2 + 1)) (x ^ 3 + x-7 ) \ right) ^ (x) = $$ $$ = \ lim \ limits_ (x \ to \ infty) \ left (1+ \ frac (2x ^ 2-x + 8) (x ^ 3 + x-7) \ right) ^ (x) = \ 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)) = $$
The expression in brackets is actually the second wonderful limit $ \ lim \ limits_ (t \ to \ infty) \ left (1+ \ frac (1) (t) \ right) ^ (t) = e $, only $ t = \ frac (x ^ 3 + x-7) (2x ^ 2-x + 8) \ to \ infty $, so
$$ = \ lim \ limits_ (x \ to \ infty) \ left (e \ right) ^ (x \ frac (2x ^ 2-x + 8) (x ^ 3 + x-7)) = \ lim \ limits_ (x \ to \ infty) e ^ (\ frac (2x ^ 2-x + 8) (x ^ 2 + 1-7 / x)) = \ lim \ limits_ (x \ to \ infty) e ^ (\ frac (2-1 / x + 8 / x ^ 2) (1 + 1 / x ^ 2-7 / x ^ 3)) = e ^ (2). $$
There are several wonderful limits, but the most famous are the first and second wonderful limits. The remarkable thing about these limits lies in the fact that they are widely used and with their help one can find other limits that are encountered in numerous problems. This is what we will do in the practical part of this lesson. To solve problems by reducing them to the first or second remarkable limit, it is not necessary to disclose the uncertainties contained in them, since the values of these limits have long been deduced by great mathematicians.
The first wonderful limit is called the limit of the ratio of the sine of an infinitesimal arc to the same arc, expressed in radian measure:
Let's move on to solving problems at the first remarkable limit. Note: if a trigonometric function is under the limit sign, this is almost a sure sign that this expression can be reduced to the first remarkable limit.
Example 1. Find the limit.
Solution. Substitution instead of x zero leads to uncertainty:
.
The denominator is sine, therefore, the expression can be reduced to the first remarkable limit. We start transformations:
.
The denominator contains the sine of three x, and the numerator has only one x, which means that you need to get three x in the numerator as well. For what? To represent 3 x = a and get an expression.
And we arrive at a variation on the first wonderful limit:
because it doesn't matter which letter (variable) in this formula stands for the x.
We multiply x by three and immediately divide:
.
In accordance with the observed first remarkable limit, we replace the fractional expression:
Now we can finally solve this limit:
.
Example 2. Find the limit.
Solution. Direct substitution again results in a zero-divide-by-zero ambiguity:
.
To get the first remarkable limit, you need the x under the sine sign in the numerator and just the x in the denominator with the same coefficient. Let this coefficient be equal to 2. To do this, we represent the current coefficient at x as below, performing actions with fractions, we get:
.
Example 3. Find the limit.
Solution. When substituting, we again obtain the "zero divided by zero" uncertainty:
.
You probably already understand that from the original expression you can get the first wonderful limit multiplied by the first wonderful limit. To do this, we decompose the squares of the x in the numerator and the sine in the denominator by the same factors, and to get the same coefficients for the x and sine, divide the x in the numerator by 3 and then multiply by 3. We get:
.
Example 4. Find the limit.
Solution. Again we get the uncertainty "zero divided by zero":
.
We can get the ratio of the first two remarkable limits. Divide both the numerator and the denominator by x. Then, so that the coefficients for sines and for x coincide, the upper x is multiplied by 2 and immediately divided by 2, and the lower x is multiplied by 3 and immediately divided by 3. We get:
Example 5. Find the limit.
Solution. And again the uncertainty "zero divided by zero":
Remember from trigonometry that the tangent is the ratio of the sine to the cosine, and the cosine of zero is one. We make transformations and get:
.
Example 6. Find the limit.
Solution. The trigonometric function under the limit sign again suggests the idea of applying the first remarkable limit. We represent it as the ratio of sine to cosine.
Find wonderful limits it is difficult not only for many first and second year students who study the theory of limits, but also for some teachers.
Formula for the first wonderful limit
Consequences of the first remarkable limit
can be written by the formulas
1. 2. 3. 4. But by themselves, the general formulas of remarkable limits do not help anyone on an exam or test. The bottom line is that real tasks are built so that you still need to come to the formulas written above. And most of the students who miss the pairs, study this course by correspondence or have teachers who themselves do not always understand what they are explaining, cannot calculate the most elementary examples to remarkable limits. From the formulas of the first remarkable limit, we see that with their help it is possible to investigate uncertainties of the type zero divided by zero for expressions with trigonometric functions. Let's first look at a number of examples for the first remarkable limit, and then explore the second remarkable limit.
Example 1. Find the limit of the function sin (7 * x) / (5 * x)
Solution: As you can see, the function under the limit is close to the first remarkable limit, but the limit of the function itself is definitely not equal to one. In this kind of tasks for the limits, you should select the variable in the denominator with the same coefficient that is contained in the variable under the sine. In this case, divide and multiply by 7 
To some, such detailing will seem superfluous, but for most students who find it difficult to give limits, it will help to better understand the rules and learn theoretical material.
Also, if there is a reverse kind of function, this is also the first remarkable limit. And all because the wonderful limit is equal to one
The same rule applies to consequences of 1 remarkable limit. Therefore, if you are asked "What is the first wonderful limit?" You must answer without hesitation that it is one.
Example 2. Find the limit of the function sin (6x) / tan (11x)
Solution: To understand the final result, let's write the function in the form 
To apply the rules of the remarkable limit, we multiply and divide by factors
Further, we write the limit of the product of functions in terms of the product of the limits 
Without complicated formulas, we have found the limit of the hour trigonometric functions... For assimilation simple formulas try to come up with and find the limit on 2 and 4 the formula of Corollary 1 of the remarkable limit. We'll look at more complex tasks.
Example 3. Calculate the limit (1-cos (x)) / x ^ 2
Solution: When checking by substitution, we get an uncertainty of 0/0. Many people do not know how to reduce such an example to one remarkable limit. The trigonometric formula should be used here 
In this case, the limit will be transformed to an understandable form 
We have succeeded in reducing the function to the square of a remarkable limit.
Example 4. Find the limit
Solution: Substitution gives the familiar 0/0 feature. However, the variable tends to Pi, not zero. Therefore, to apply the first remarkable limit, we change the variable x so that the new variable goes to zero. To do this, denote the denominator as a new variable Pi-x = y 
Thus, using the trigonometric formula, which is given in the previous task, the example is reduced to 1 remarkable limit.
Example 5. Calculate the limit 
Solution: At first, it is not clear how to simplify the limits. But since there is an example, then there must be an answer. The fact that the variable is directed to one gives, when substituted, a singularity of the form zero multiplied by infinity, therefore the tangent must be replaced by the formula 
After that, we get the desired uncertainty 0/0. Next, we perform the change of variables in the limit, and use the periodicity of the cotangent 
The last substitutions allow us to use Corollary 1 of the remarkable limit.
The second remarkable limit is exponential
This is a classic to which in real problems it is not always easy to reach the limits.
In calculations you will need limits are consequences of the second remarkable limit:
1. 
2. 
3. 
4.
Thanks to the second remarkable limit and its consequences, it is possible to study uncertainties such as zero divided by zero, one to the power of infinity, and infinity divided by infinity, and even to the same degree. 
Let's start to get familiar with simple examples.
Example 6. Find the limit of a function
Solution: You cannot directly apply the 2 wonderful limit. First, you should turn the exponent so that it has the opposite form to the term in parentheses 
This is the technique of reducing to the 2 remarkable limit and, in fact, the derivation 2 of the formula for the corollary of the limit.
Example 7. Find the limit of a function
Solution: We have tasks for 3 the formula of Corollary 2 of a remarkable limit. Substitution of zero gives a feature of the form 0/0. To raise the limit under the rule, turn the denominator so that the variable has the same coefficient as in the logarithm 
It is also easy to understand and complete on the exam. Students' difficulty in calculating limits begins with the following tasks.
Example 8. Calculate the limit of a function[(x + 7) / (x-3)] ^ (x-2) 
Solution: We have a singularity of type 1 to the degree of infinity. If you do not believe, you can substitute infinity instead of "X" everywhere and make sure of it. To build under the rule, divide the numerator by the denominator in parentheses, for this we first perform the manipulations 
Substitute the expression into the limit and turn to 2 the remarkable limit 
The limit is exponential to the 10th power. Constants, which are addends for a variable both in brackets and a degree, do not bring any "weather" - this should be remembered. And if the teachers ask you - "Why don't you turn the indicator?" (For this example in x-3), then say that "When a variable tends to infinity, then at least add 100 to it, or subtract 1000, and the limit will remain the same!".
There is a second way to calculate these types of limits. We will talk about it in the next task.
Example 9. Find the limit 
Solution: Now let's move out the variable in the numerator and denominator and convert one feature to another. To obtain the final value, we use the formula of Corollary 2 of the remarkable limit 
Example 10. Find the limit of a function
Solution: Not everyone can find the set limit. To raise the limit under 2, imagine that sin (3x) is a variable, but you need to turn the exponent 
Further, we write the indicator as a degree to a degree 
Intermediate reasoning is described in parentheses. As a result of using the first and second remarkable limit, we got an exponent in a cube.
Example 11. Calculate the limit of a function sin (2 * x) / ln (3 * x + 1)
Solution: We have an uncertainty of the form 0/0. In addition, we see that the function should be converted to use both remarkable limits. Let's perform the previous mathematical transformations 
Further, without difficulty, the limit will take on the value 
This is how you will feel at ease on examinations, tests, modules if you learn to quickly describe functions and bring them under the first or second wonderful limit. If it is difficult for you to memorize the given methods of finding the limits, then you can always order test to our limits.
To do this, fill out the form, specify the data and attach a file with examples. We have helped many students - we can help you too!
This article: "The second remarkable limit" is devoted to disclosure within the uncertainties of the form:
$ \ bigg [\ frac (\ infty) (\ infty) \ bigg] ^ \ infty $ and $ ^ \ infty $.
Also, such uncertainties can be disclosed using the logarithm of the exponential function, but this is a different solution method, which will be covered in another article.
Formula and Consequences
Formula the second remarkable limit is written as follows: $$ \ lim_ (x \ to \ infty) \ bigg (1+ \ frac (1) (x) \ bigg) ^ x = e, \ text (where) e \ approx 2.718 $$
The formula implies consequences, which are very convenient to use for solving examples with limits: $$ \ lim_ (x \ to \ infty) \ bigg (1 + \ frac (k) (x) \ bigg) ^ x = e ^ k, \ text (where) k \ in \ mathbb (R) $$ $$ \ lim_ (x \ to \ infty) \ bigg (1 + \ frac (1) (f (x)) \ bigg) ^ (f (x)) = e $ $ $$ \ lim_ (x \ to 0) \ bigg (1 + x \ bigg) ^ \ frac (1) (x) = e $$
It is worth noting that the second remarkable limit can be applied not always to the exponential function, but only in cases where the base tends to unity. To do this, first, the base limit is calculated in the mind, and then conclusions are drawn. All of this will be covered in the sample solutions.
Examples of solutions
Let's consider examples of solutions using a direct formula and its consequences. We will also analyze the cases in which the formula is not needed. It is enough to write down only the ready-made answer.
| Example 1 |
| Find the limit $ \ lim_ (x \ to \ infty) \ bigg (\ frac (x + 4) (x + 3) \ bigg) ^ (x + 3) $ |
| Solution |
|
Let's substitute infinity in the limit and look at the uncertainty: $$ \ lim_ (x \ to \ infty) \ bigg (\ frac (x + 4) (x + 3) \ bigg) ^ (x + 3) = \ bigg (\ frac (\ infty) (\ infty) \ bigg) ^ \ infty $$ Find the limit of the base: $$ \ lim_ (x \ to \ infty) \ frac (x + 4) (x + 3) = \ lim_ (x \ to \ infty) \ frac (x (1+ \ frac (4) ( x))) (x (1+ \ frac (3) (x))) = 1 $$ We got a base equal to one, which means that the second remarkable limit can already be applied. To do this, we fit the base of the function to the formula by subtracting and adding one: $$ \ lim_ (x \ to \ infty) \ bigg (1 + \ frac (x + 4) (x + 3) - 1 \ bigg) ^ (x + 3) = \ lim_ (x \ to \ infty) \ bigg (1 + \ frac (1) (x + 3) \ bigg) ^ (x + 3) = $$ We look at the second consequence and write down the answer: $$ \ lim_ (x \ to \ infty) \ bigg (1 + \ frac (1) (x + 3) \ bigg) ^ (x + 3) = e $$ If you can't solve your problem, then send it to us. We will provide detailed solution... You will be able to familiarize yourself with the course of the calculation and get information. This will help you get credit from your teacher in a timely manner! |
| Answer |
| $$ \ lim_ (x \ to \ infty) \ bigg (1 + \ frac (1) (x + 3) \ bigg) ^ (x + 3) = e $$ |
| Example 4 |
| Solve limit $ \ lim_ (x \ to \ infty) \ bigg (\ frac (3x ^ 2 + 4) (3x ^ 2-2) \ bigg) ^ (3x) $ |
| Solution |
|
We find the limit of the base and see that $ \ lim_ (x \ to \ infty) \ frac (3x ^ 2 + 4) (3x ^ 2-2) = 1 $, so the second wonderful limit can be applied. Standardly, according to the plan, we add and subtract one from the base of the degree: $$ \ lim_ (x \ to \ infty) \ bigg (1+ \ frac (3x ^ 2 + 4) (3x ^ 2-2) -1 \ bigg) ^ (3x) = \ lim_ (x \ to \ infty ) \ bigg (1+ \ frac (6) (3x ^ 2-2) \ bigg) ^ (3x) = $$ We fit the fraction to the formula of the 2nd remark. limit: $$ = \ lim_ (x \ to \ infty) \ bigg (1+ \ frac (1) (\ frac (3x ^ 2-2) (6)) \ bigg) ^ (3x) = $$ Now we adjust the degree. The power must be a fraction equal to the denominator of the base $ \ frac (3x ^ 2-2) (6) $. To do this, multiply and divide the degree by it, and continue to solve: $$ = \ lim_ (x \ to \ infty) \ bigg (1+ \ frac (1) (\ frac (3x ^ 2-2) (6)) \ bigg) ^ (\ frac (3x ^ 2-2) (6) \ cdot \ frac (6) (3x ^ 2-2) \ cdot 3x) = \ lim_ (x \ to \ infty) e ^ (\ frac (18x) (3x ^ 2-2)) = $$ The limit located in degree at $ e $ is: $ \ lim_ (x \ to \ infty) \ frac (18x) (3x ^ 2-2) = 0 $. Therefore, continuing the solution, we have: |
| Answer |
| $$ \ lim_ (x \ to \ infty) \ bigg (\ frac (3x ^ 2 + 4) (3x ^ 2-2) \ bigg) ^ (3x) = 1 $$ |
Let us examine the cases when the problem is similar to the second remarkable limit, but it can be solved without it.
In the article: "The second remarkable limit: examples of solutions" was analyzed the formula, its consequences, and given the frequent types of problems on this topic.
Proof:
First, we prove the theorem for the case of the sequence 
According to the binomial Newton formula:
Assuming we get
From this equality (1) it follows that as n increases, the number of positive terms on the right-hand side increases. In addition, as n increases, the number decreases, so the quantities 
increase. Therefore, the sequence 
increasing, while (2) * Let us show that it is bounded. Replace each parenthesis on the right side of the equality with one, right part increases, we obtain the inequality
Let us strengthen the resulting inequality, replace 3,4,5, ..., standing in the denominators of fractions, with the number 2: 
(3)*
So, the sequence is bounded from above, while inequalities (2) and (3) hold: 
Therefore, based on the Weierstrass theorem (a criterion for the convergence of a sequence), the sequence 
monotonically increasing and limited, which means it has a limit, denoted by the letter e. Those. 
Knowing that the second remarkable limit is true for natural values of x, we prove the second remarkable limit for real x, that is, we will prove that 
... Consider two cases:
1. Let Each value of x be enclosed between two positive integers:, where is whole part x. => =>
If, then Therefore, according to the limit 
We have
On the basis (about the limit of the intermediate function) of the existence of the limits 
2. Let. We make the substitution - x = t, then
From these two cases it follows that 
for real x.
Consequences:
9 .) Comparison of infinitesimal. The theorem on the replacement of infinitesimal by equivalent in the limit and the theorem on the principal part of the infinitesimal.
Let the functions a ( x) and b ( x) - b.m. at x ® x 0 .
DEFINITIONS.
1) a ( x) called infinitesimal of higher order than b (x) if
Write: a ( x) = o (b ( x)) .
2) a ( x) and b ( x)are called infinitesimal of the same order, if
where CÎℝ and C¹ 0 .
Write: a ( x) = O(b ( x)) .
3) a ( x) and b ( x) are called equivalent , if
Write: a ( x) ~ b ( x).
4) a ( x) is called infinitesimal of order k with respect to
infinitely small b ( x),
if infinitesimal a ( x)and(b ( x)) k are of the same order, i.e. if
where CÎℝ and C¹ 0 .
THEOREM 6 (on the replacement of infinitesimal by equivalent ones).
Let be a ( x), b ( x), a 1 ( x), b 1 ( x)- b.m. at x ® x 0 ... If a ( x) ~ a 1 ( x), b ( x) ~ b 1 ( x),
then 
Proof: Let a ( x) ~ a 1 ( x), b ( x) ~ b 1 ( x), then
THEOREM 7 (about the main part of the infinitesimal).
Let be a ( x)and b ( x)- b.m. at x ® x 0 , and b ( x)- b.m. higher order than a ( x).
=, a since b ( x) - of a higher order than a ( x), then, i.e. 
from it is clear that a ( x) + b ( x) ~ a ( x)
10) Continuity of a function at a point (in the language of epsilon-delta limits, geometric) One-sided continuity. Continuity on an interval, on a segment. Properties of continuous functions.
1. Basic definitions
Let be f(x) is defined in some neighborhood of the point x 0 .
DEFINITION 1. Function f(x) called continuous at point x 0 if the equality is true
Remarks.
1) By virtue of Theorem 5 §3, equality (1) can be written in the form
Condition (2) - definition of the continuity of a function at a point in the language of one-sided limits.
2) Equality (1) can also be written as:
They say: “if the function is continuous at the point x 0, then the sign of the limit and the function can be reversed. "
DEFINITION 2 (in language e-d).
Function f(x) called continuous at point x 0 if"e> 0 $ d> 0 such, what
if xÎU ( x 0, d) (i.e. | x – x 0 | < d),
then 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)
We denote: D x= x - x 0 – argument increment
D f(x 0) = f(x) – f(x 0) – function increment at point x 0
DEFINITION 3 (geometric).
Function f(x) on called continuous at point
x 0
if at this point the infinitesimal increment of the argument corresponds to the infinitesimal increment of the function, i.e.
Let the function f(x) is defined 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 the equality is true
It's obvious that f(x) is continuous at the point x 0 Û f(x) is continuous at the point x 0 right and left.
DEFINITION. Function f(x) called continuous for an interval e ( a; b) if it is continuous at every point of this interval.
Function f(x) is called continuous on the segment [a; b] if it is continuous on the interval (a; b) and has one-sided continuity at the boundary points(i.e., continuous at the point a on the right, at the point b- left).
11) Break points, their classification
DEFINITION. If the function f(x) defined in some neighborhood of the point x 0 , but is not continuous at this point, then f(x) is called discontinuous at the point x 0 , but the point itself x 0 called a break point function f(x) .
Remarks.
1) f(x) can be defined in an incomplete neighborhood of the point x 0 .
Then the corresponding one-sided continuity of the function is considered.
2) From the definition of Þ point x 0 is the discontinuity point of the function f(x) in two cases:
a) U ( x 0, d) Î D(f) , but for f(x) the equality
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 break point I kind if the function f(x)has finite limits on the left and right at this point.
If, in addition, these limits are equal, then the point x 0 called point of removable discontinuity , otherwise - jump point .
DEFINITION. Point x 0 called break point II kind if at least one of the one-sided limits of the function f(x)at this point is¥ or does not exist.
12) Properties of functions continuous on an interval (theorems of Weierstrass (without proof) and Cauchy
Weierstrass theorem
Let the function f (x) be continuous on an interval, then
1) f (x) is bounded on
2) f (x) takes its smallest and largest value on the interval
Definition: The value of the function m = f is called the smallest if m≤f (x) for any x € D (f).
The value of the function m = f is called the largest if m≥f (x) for any x ∈ D (f).
The function can take the smallest \ largest value at several points of the segment.
f (x 3) = f (x 4) = max
Cauchy's theorem.
Let the function f (x) be continuous on an interval and x is a number between f (a) and f (b), then there exists at least one point x 0 € such that f (x 0) = g
