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 remarkable limit
Consequences of the first remarkable limit
let's write it in formulas
1. 2. 3. 4. But on their own general formulas remarkable limits do not help anyone in an exam or test. The point is that real tasks are constructed so that you still need to arrive at the formulas written above. And the majority of students who miss classes, study this course in absentia, 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 study uncertainties of the type zero divided by zero for expressions with trigonometric functions. Let us first consider a number of examples for the first wonderful limit y, and then we will study 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 on limits, one should select in the denominator a variable with the same coefficient as that contained in the variable under the sine. In this case, divide and multiply by 7 
For some, such detail will seem unnecessary, but for most students who find it difficult to understand limits, it will help them better understand the rules and learn theoretical material.
Also, if there is an inverse form of a function, this is also the first wonderful limit. And all because the wonderful limit is equal to one
The same rule applies to the consequences of the 1st remarkable limit. Therefore, if you are asked, “What is the first remarkable limit?” You should answer without hesitation that it is a unit.
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, multiply and divide by factors
Next, we write the limit of a product of functions through the product of limits 
Without complex formulas, we found the chaska limit trigonometric functions. To master simple formulas, try to come up with and find the limit on 2 and 4, the formula for the corollary of 1 wonderful limit. We will look at more complex problems.
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. Here you should use trigonometric formula 
In this case, the limit will transform to a clear form 
We managed to reduce the function to the square of a remarkable limit.
Example 4. Find the limit
Solution: When substituting, we get the familiar feature 0/0. However, the variable tends to Pi rather than zero. Therefore, to apply the first remarkable limit, we will perform such a change in the variable x so that the new variable goes to zero. To do this, we denote the denominator as a new variable Pi-x=y 
Thus, using the trigonometric formula given in the previous task, the example is reduced to 1 remarkable limit.
Example 5: Calculate 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 goes to unity gives, when substituting, a feature of the form zero multiplied by infinity, so the tangent must be replaced using the formula 
After this we get the required uncertainty 0/0. Next, we perform a 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 equal to the exponential
This is a classic that is not always easy to reach in real limit problems.
In the 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 explore 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 with simple examples.
Example 6. Find the limit of a function
Solution: Directly applying the 2nd remarkable limit will not work. First, you should transform the exponent so that it looks like the inverse of the term in brackets 
This is the technique of reducing to the 2nd remarkable limit and, in essence, deducing the 2nd formula for the corollary of the limit.
Example 7. Find the limit of a function
Solution: We have tasks for formula 3 of corollary 2 of a wonderful limit. Substituting zero gives a singularity of the form 0/0. To raise the limit to a rule, we turn the denominator so that the variable has the same coefficient as in the logarithm 
It is also easy to understand and perform in the exam. Students' difficulties in calculating limits begin with the following problems.
Example 8. Calculate the limit of a function[(x+7)/(x-3)]^(x-2) 
Solution: We have a type 1 singularity to the power of infinity. If you don’t believe me, you can substitute infinity for “X” everywhere and make sure of it. To construct a rule, we divide the numerator by the denominator in parentheses; to do this, we first perform the manipulations 
Let's substitute the expression into the limit and turn it into 2 wonderful limit 
The limit is equal to the exponential power of 10. Constants that are terms with a variable, both in parentheses and a degree, do not introduce any “weather” - this should be remembered. And if your teachers ask you, “Why don’t you convert the indicator?” (For this example in x-3), then say that “When a variable tends to infinity, then even add 100 to it or subtract 1000, and the limit will remain the same as it was!”
There is a second way to calculate limits of this type. We'll talk about it in the next task.
Example 9. Find the limit 
Solution: Now let's take out the variable in the numerator and denominator and turn one feature into 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 given limit. To raise the limit to 2, imagine that sin (3x) is a variable, and you need to turn the exponent 
Next, we write the indicator as a power to a power 
Intermediate arguments are described in parentheses. As a result of using the first and second remarkable limits, we obtained the exponential in 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 wonderful limits. Let's perform the previous mathematical transformations 
Further, without difficulty, the limit will take the value 
This is how free you will feel on assignments, tests, modules if you learn to quickly write out functions and reduce them to the first or second wonderful limit. If it is difficult for you to memorize the given methods for finding limits, then you can always order a test paper on limits from us.
To do this, fill out the form, provide data and attach a file with examples. We have helped many students - we can help you too!
There are several remarkable limits, but the most famous are the first and second remarkable limits. The remarkable thing about these limits is that they are widely used and with their help one can find other limits 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, there is no need to reveal 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 there is a trigonometric function under the limit sign, this is almost sure sign that this expression can be taken to its first remarkable limit.
Example 1. Find the limit.
Solution. Substitution instead x zero leads to uncertainty:
.
The denominator is sine, therefore, the expression can be brought to the first remarkable limit. Let's start the transformation:
.
The denominator is the sine of three X, but the numerator has only one X, which means you need to get three X in the numerator. For what? To introduce 3 x = a and get the expression .
And we come to a variation of the first remarkable limit:
because it doesn’t matter which letter (variable) in this formula stands instead of X.
We multiply X by three and immediately divide:
.
In accordance with the first remarkable limit noticed, we replace the fractional expression:
Now we can finally solve this limit:
.
Example 2. Find the limit.
Solution. Direct substitution again leads to the “zero divided by zero” uncertainty:
.
To get the first remarkable limit, it is necessary that the x under the sine sign in the numerator and just the x in the denominator have the same coefficient. Let this coefficient be equal to 2. To do this, imagine the current coefficient for x as below, performing operations with fractions, we obtain:
.
Example 3. Find the limit.
Solution. When substituting, we again get the uncertainty “zero divided by zero”:
.
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 into identical factors, and in order to get the same coefficients for the x and sine, we divide the x in the numerator by 3 and immediately multiply by 3. We get:
.
Example 4. Find the limit.
Solution. Once again we get the uncertainty “zero divided by zero”:
.
We can obtain the ratio of the first two remarkable limits. We divide both the numerator and the denominator by x. Then, so that the coefficients for sines and xes coincide, we multiply the upper x by 2 and immediately divide by 2, and multiply the lower x by 3 and immediately divide by 3. We get:
Example 5. Find the limit.
Solution. And again the uncertainty of “zero divided by zero”:
We remember from trigonometry that tangent is the ratio of sine to cosine, and the cosine of zero is equal to one. We carry out the transformations and get:
.
Example 6. Find the limit.
Solution. The trigonometric function under the sign of a limit again suggests the use of the first remarkable limit. We represent it as the ratio of sine to cosine.
The formula for the second remarkable limit is lim x → ∞ 1 + 1 x x = e. Another form of writing looks like this: lim x → 0 (1 + x) 1 x = e.
When we talk about the second remarkable limit, we have to deal with uncertainty of the form 1 ∞, i.e. unity to an infinite degree.
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Let's consider problems in which the ability to calculate the second remarkable limit will be useful.
Example 1
Find the limit lim x → ∞ 1 - 2 x 2 + 1 x 2 + 1 4 .
Solution
Let's substitute the required formula and perform the calculations.
lim x → ∞ 1 - 2 x 2 + 1 x 2 + 1 4 = 1 - 2 ∞ 2 + 1 ∞ 2 + 1 4 = 1 - 0 ∞ = 1 ∞
Our answer turned out to be one to the power of infinity. To determine the solution method, we use the uncertainty table. Let's choose the second remarkable limit and make a change of variables.
t = - x 2 + 1 2 ⇔ x 2 + 1 4 = - t 2
If x → ∞, then t → - ∞.
Let's see what we got after the replacement:
lim x → ∞ 1 - 2 x 2 + 1 x 2 + 1 4 = 1 ∞ = lim x → ∞ 1 + 1 t - 1 2 t = lim t → ∞ 1 + 1 t t - 1 2 = e - 1 2
Answer: lim x → ∞ 1 - 2 x 2 + 1 x 2 + 1 4 = e - 1 2 .
Example 2
Calculate the limit lim x → ∞ x - 1 x + 1 x .
Solution
Let's substitute infinity and get the following.
lim x → ∞ x - 1 x + 1 x = lim x → ∞ 1 - 1 x 1 + 1 x x = 1 - 0 1 + 0 ∞ = 1 ∞
In the answer, we again got the same thing as in the previous problem, therefore, we can again use the second remarkable limit. Next we need to select at the base power function whole part:
x - 1 x + 1 = x + 1 - 2 x + 1 = x + 1 x + 1 - 2 x + 1 = 1 - 2 x + 1
After this, the limit takes on the following form:
lim x → ∞ x - 1 x + 1 x = 1 ∞ = lim x → ∞ 1 - 2 x + 1 x
Replace variables. Let's assume that t = - x + 1 2 ⇒ 2 t = - x - 1 ⇒ x = - 2 t - 1 ; if x → ∞, then t → ∞.
After that, we write down what we got in the original limit:
lim x → ∞ x - 1 x + 1 x = 1 ∞ = lim x → ∞ 1 - 2 x + 1 x = lim x → ∞ 1 + 1 t - 2 t - 1 = = lim x → ∞ 1 + 1 t - 2 t 1 + 1 t - 1 = lim x → ∞ 1 + 1 t - 2 t lim x → ∞ 1 + 1 t - 1 = = lim x → ∞ 1 + 1 t t - 2 1 + 1 ∞ = e - 2 · (1 + 0) - 1 = e - 2
To perform this transformation, we used the basic properties of limits and powers.
Answer: lim x → ∞ x - 1 x + 1 x = 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 .
Solution
lim x → ∞ x 3 + 1 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 = lim x → ∞ 1 + 1 x 3 1 + 2 x - 1 x 3 3 2 x - 5 x 4 = = 1 + 0 1 + 0 - 0 3 0 - 0 = 1 ∞
After that, we need to transform the function to apply the second great limit. We got the following:
lim x → ∞ x 3 + 1 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 = 1 ∞ = 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 = = 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 = 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 = = 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 we now have the same exponents in the numerator and denominator of the fraction (equal to six), the limit of the fraction at infinity will be equal to the ratio of these coefficients at higher powers.
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 = = 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 = lim x → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 x 3 + 2 x 2 - 1 - 2 x 2 + 2 - 3
By substituting t = x 2 + 2 x 2 - 1 - 2 x 2 + 2 we get a second remarkable 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 = lim x → ∞ 1 + 1 t t - 3 = e - 3
Answer: lim x → ∞ x 3 + 1 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 = e - 3 .
conclusions
Uncertainty 1 ∞, i.e. unity to an infinite power is a power-law uncertainty, therefore, it can be revealed using the rules for finding the limits of exponential power functions.
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From the above article you can find out what the limit is and what it is eaten with - this is VERY important. Why? You may not understand what determinants are and successfully solve them; you may not understand at all what a derivative is and find them with an “A”. But if you don’t understand what a limit is, then solving practical tasks will be difficult. It would also be a good idea to familiarize yourself with the sample solutions and my design recommendations. All information is presented in a simple and accessible form.
And for the purposes of this lesson we will need the following teaching materials: Wonderful Limits And Trigonometric formulas. They can be found on the page. It is best to print out the manuals - it is much more convenient, and besides, you will often have to refer to them offline.
What is so special about remarkable limits? The remarkable thing about these limits is that they were proven by the greatest minds of famous mathematicians, and grateful descendants do not have to suffer from terrible limits with a pile of trigonometric functions, logarithms, powers. That is, when finding the limits, we will use ready-made results that have been proven theoretically.
There are several wonderful limits, but in practice, in 95% of cases, part-time students have two wonderful limits: The first wonderful limit, Second wonderful limit. It should be noted that these are historically established names, and when, for example, they talk about “the first remarkable limit,” they mean by this a very specific thing, and not some random limit taken from the ceiling.
The first wonderful limit
Consider the following limit: (instead of the native letter “he” I will use the Greek letter “alpha”, this is more convenient from the point of view of presenting the material).
According to our rule for finding limits (see article Limits. Examples of solutions) we try to substitute zero into the function: in the numerator we get zero (the sine of zero is zero), and in the denominator, obviously, there is also zero. Thus, we are faced with an uncertainty of the form, which, fortunately, does not need to be disclosed. I know mathematical analysis, it is proved that:
This mathematical fact is called The first wonderful limit. I won’t give an analytical proof of the limit, but here it is: geometric meaning we'll look at it in class about infinitesimal functions.
Often in practical tasks functions can be arranged differently, this does not change anything:
- the same first wonderful limit.
But you cannot rearrange the numerator and denominator yourself! If a limit is given in the form , then it must be solved in the same form, without rearranging anything.
In practice, not only a variable can act as a parameter, but also elementary function, complex function. The only important thing is that it tends to zero.
Examples:
, , 
, 
Here , , , 
, and everything is good - the first wonderful limit is applicable.
But the following entry is heresy:
Why? Because the polynomial does not tend to zero, it tends to five.
By the way, a quick question: what is the limit? 
? The answer can be found at the end of the lesson.
In practice, not everything is so smooth; almost never a student is offered to solve a free limit and get an easy pass. Hmmm... I’m writing these lines, and a very important thought came to mind - after all, it’s better to remember “free” mathematical definitions and formulas by heart, this can provide invaluable help in the test, when the question will be decided between “two” and “three”, and the teacher decides to ask the student some simple question or offer to solve simplest example(“maybe he (s) still knows what?!”).
Let's move on to consider practical examples:
Example 1
Find the limit
If we notice a sine in the limit, then this should immediately lead us to think about the possibility of applying the first remarkable limit.
First, we try to substitute 0 into the expression under the limit sign (we do this mentally or in a draft):
So we have an uncertainty of the form be sure to indicate in making a decision. The expression under the limit sign is similar to the first wonderful limit, but this is not exactly it, it is under the sine, but in the denominator.
In such cases, we need to organize the first remarkable limit ourselves, using an artificial technique. The line of reasoning could be as follows: “under the sine we have , which means that we also need to get in the denominator.”
And this is done very simply:
That is, the denominator is artificially multiplied in this case by 7 and divided by the same seven. Now our recording has taken on a familiar shape.
When the task is drawn up by hand, it is advisable to mark the first remarkable limit with a simple pencil:
What happened? In fact, our circled expression turned into a unit and disappeared in the work: 
Now all that remains is to get rid of the three-story fraction: 
Who has forgotten the simplification of multi-level fractions, please refresh the material in the reference book Hot formulas for school mathematics course .
Ready. Final answer:
If you don’t want to use pencil marks, then the solution can be written like this:
“
Let's use the first wonderful limit 
“
Example 2
Find the limit
Again we see a fraction and a sine in the limit. Let’s try to substitute zero into the numerator and denominator:
Indeed, we have uncertainty and, therefore, we 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, we need to factorize the numerator and denominator. Here it’s the same thing, we’ll represent the degrees as a product (multipliers):
Similar to the previous example, we draw a pencil around the remarkable limits (here there are two of them), and indicate that they tend to unity:
Actually, the answer is ready:
In the following examples, I will not do art in Paint, I think how to correctly draw up a solution in a notebook - you already understand.
Example 3
Find the limit
We substitute zero into the expression under the limit sign:
An uncertainty has been obtained that needs to be disclosed. If there is a tangent in the limit, then it is almost always converted into sine and cosine using the well-known trigonometric formula (by the way, they do approximately the same thing with cotangent, see Fig. methodological material Hot trigonometric formulas On the page Mathematical formulas, tables and reference materials).
In this case:
The cosine of zero is equal to one, and it’s easy to get rid of it (don’t forget to mark that it tends to one):
Thus, if in the limit the cosine is a MULTIPLIER, then, roughly speaking, it needs to be turned into a unit, which disappears in the product.
Here everything turned out simpler, without any multiplications and divisions. The first remarkable limit also turns into one and disappears in the product:
As a result, infinity is obtained, and this happens.
Example 4
Find the limit
Let's try to substitute zero into the numerator and denominator:
The uncertainty is obtained (the cosine of zero, as we remember, is equal to one)
We use the trigonometric formula. Take note! For some reason, limits using this formula are very common.
Let us move the constant factors beyond the limit icon:
Let's organize the first wonderful limit:
Here we have only one remarkable limit, which turns into one and disappears in the product:
Let's get rid of the three-story structure:
The limit is actually solved, we indicate that the remaining sine tends to zero:
Example 5
Find the limit 
This example is more complicated, try to figure it out yourself:
Some limits can be reduced to the 1st remarkable limit by changing a variable, you can read about this a little later in the article Methods for solving limits.
Second wonderful limit
In the theory of mathematical analysis it has been proven that:
This fact is called second wonderful limit.
Reference: 
is an irrational number.
The parameter can be not only a variable, but also a complex function. The only important thing is that it strives for infinity.
Example 6
Find the limit
When the expression under the limit sign is in 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 endlessly big number in the expression on what principle this is done, discussed in the lesson Limits. Examples of solutions.
It is easy to notice that when the base of the degree is , and the exponent is , that is, there is uncertainty of the form:
This uncertainty is precisely revealed with the help of the second remarkable limit. But, as often happens, the second wonderful limit does not lie on a silver platter, and it needs to be artificially organized. One can reason as follows: in in this example parameter, which means that in the indicator we also need to organize . To do this, we raise the base to the power, and so that the expression does not change, we raise it to the power:
When the task is completed by hand, we mark with a pencil:
Almost everything is ready, the terrible degree has turned into a nice letter:
In this case, we move the limit icon itself to the indicator:
Example 7
Find the limit
Attention! This type of limit occurs very often, please study this example very carefully.
Let's try to substitute an infinitely large number into the expression under the limit sign:
The result is uncertainty. But the second remarkable limit applies to the uncertainty of the form. What to do? We need to convert the base of the degree. We reason like this: in the denominator we have , which means that in the numerator we also need to organize .