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 you can find other limits that are found 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 the limit of the ratio of the sine of an infinitesimal arc to the same arc, expressed in radian measure:

We proceed to solving problems for the first wonderful 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. Let's 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) is in this formula instead of the x.

We multiply x by three and then 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 leads to the 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 x squares 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, 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 "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 equal to one. We make transformations and get:

.

Example 6. Find the limit.

Solution. The trigonometric function under the limit sign again suggests the idea of ​​using 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 notation looks like this: lim x → 0 (1 + x) 1 x = e.

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

Yandex.RTB R-A-339285-1

Consider problems in which the ability to calculate the second remarkable limit will come in handy.

Example 1

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

Solution

Substitute the desired 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 ∞

In our answer, we got one to the power of infinity. To determine the solution method, we use the table of uncertainties. 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

Compute the limit lim x → ∞ x - 1 x + 1 x.

Solution

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 that, the limit takes the following form:

lim x → ∞ x - 1 x + 1 x = 1 ∞ = lim x → ∞ 1 - 2 x + 1 x

We replace the variables. Suppose 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 tt - 2 1 + 1 ∞ = e - 2 (1 + 0) - 1 = e - 2

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

Answer: lim x → ∞ x - 1 x + 1 x = e - 2.

Example 3

Find 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 remarkable 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 now we 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 the highest 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

Replacing t = x 2 + 2 x 2 - 1 - 2 x 2 + 2 gives us 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 tt - 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. unit to an infinite degree is a power uncertainty, therefore, it can be expanded using the rules for finding the limits of exponential functions.

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A bit of theory.

Limit of the function at x-> x 0

Let the function f (x) be defined on some set X and let the point \ (x_0 \ in X \) or \ (x_0 \ notin X \)

Take from X a sequence of points other than x 0:
x 1, x 2, x 3, ..., x n, ... (1)
converging to x *. The values ​​of the function at the points of this sequence also form a numerical sequence
f (x 1), f (x 2), f (x 3), ..., f (x n), ... (2)
and the question of the existence of its limit can be raised.

Definition... The number A is called the limit of the function f (x) at the point x = x 0 (or at x -> x 0) if for any sequence (1) converging to x 0 of the values ​​of the argument x other than x 0 the corresponding sequence (2) of values function converges to A.

$$ \ lim_ (x \ to x_0) (f (x)) = A $$

The function f (x) can have only one limit at the point x 0. This follows from the fact that the sequence
(f (x n)) has only one limit.

There is another definition of a function limit.

Definition The number A is called the limit of the function f (x) at the point x = x 0 if for any number \ (\ varepsilon> 0 \) there exists a number \ (\ delta> 0 \) such that for all \ (x \ in X, \; x \ neq x_0 \) satisfying the inequality \ (| x-x_0 | Using logical symbols, this definition can be written as
\ ((\ forall \ varepsilon> 0) (\ exists \ delta> 0) (\ forall x \ in X, \; x \ neq x_0, \; | x-x_0 | Note that the inequalities \ (x \ neq x_0 , \; | x-x_0 | The first definition is based on the notion of a limit numerical sequence therefore it is often referred to as the "sequence language" definition. The second definition is called the "\ (\ varepsilon - \ delta \)" definition.
These two definitions of the limit of a function are equivalent and you can use any of them, depending on which is more convenient for solving a particular problem.

Note that the definition of the limit of a function "in the language of sequences" is also called the definition of the limit of a function according to Heine, and the definition of the limit of a function "in the language \ (\ varepsilon - \ delta \)" is called the definition of the limit of a function according to Cauchy.

The limit of the function at x-> x 0 - and at x-> x 0 +

In what follows, we will use the concept of one-sided function limits, which are defined as follows.

Definition The number A is called the right (left) limit of the function f (x) at the point x 0 if for any sequence (1) converging to x 0, whose elements x n are greater (less) x 0, the corresponding sequence (2) converges to A.

This is symbolically written as follows:
$$ \ lim_ (x \ to x_0 +) f (x) = A \; \ left (\ lim_ (x \ to x_0-) f (x) = A \ right) $$

You can give an equivalent definition of one-way limits of a function "in the language \ (\ varepsilon - \ delta \)":

Definition the number A is called the right (left) limit of the function f (x) at the point x 0 if for any \ (\ varepsilon> 0 \) there exists \ (\ delta> 0 \) such that for all x satisfying the inequalities \ (x_0 Symbolic entries:

\ ((\ forall \ varepsilon> 0) (\ exists \ delta> 0) (\ forall x, \; x_0

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's strengthen the resulting inequality, replace 3,4,5, ..., standing in the denominators of fractions, with the number 2: The sum in parentheses is found by the formula for the sum of the terms of a geometric progression: Therefore (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 the 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 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., is 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 smallest / largest value the function can take 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

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 in the top five. But if you do not understand what the limit is, then it will be difficult to solve practical tasks. Also, it will not be superfluous to get acquainted with the samples of the design of solutions and my recommendations for design. All information is presented in a simple and accessible form.

And for the purposes of this lesson, we need the following teaching materials: Wonderful limits and Trigonometric formulas... They can be found on the page. It is best to print the manuals - it is much more convenient, and besides, they often have to be accessed offline.

Why are wonderful limits so wonderful? The remarkable thing about these limits is that they are proved greatest minds famous mathematicians, and grateful descendants do not have to suffer from terrible limits with a pile trigonometric functions, logarithms, degrees. That is, when finding the limits, we will use ready-made results that have been theoretically proven.

There are several remarkable limits, but in practice, part-time students in 95% of cases have two remarkable limits: The first wonderful limit, Second wonderful limit... It should be noted that these are historically established names, and when, for example, they speak of “the first wonderful limit,” they mean by this a very definite 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", it is more convenient from the point of view of material presentation).

According to our rule of 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), in the denominator, obviously, is also zero. Thus, we are faced with an uncertainty of the species, which, fortunately, does not need to be disclosed. In the course of mathematical analysis, it is proved that:

This mathematical fact is called The first wonderful limit... I will not give an analytical proof of the limit, but we will consider its geometric meaning in the lesson about infinitesimal functions.

Often in practical assignments functions can be arranged differently, this does not change anything:

- the same first wonderful limit.

But you cannot rearrange the numerator and denominator on your own! If the 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. It is only important that she strives for zero..

Examples:
, , ,

Here , , , and it's all good - the first wonderful limit is applicable.

But the next entry is heresy:

Why? Because the polynomial does not tend to zero, it tends to the five.

By the way, a question for filling, why equal to the limit ? The answer can be found at the end of the lesson.

In practice, not everything is so smooth, almost never a student will be offered to solve the free limit and get an easy test. Hmmm ... I am writing these lines, and a very important thought came to my mind - after all, "free" mathematical definitions and formulas seem to be better remembered by heart, this can provide invaluable help in the test, when the issue is decided between "two" and "three", and the teacher decides to ask the student some simple question or suggest solving simplest example(“Maybe he (a) still knows what ?!”).

Let's move on to consideration practical examples:

Example 1

Find the limit

If we notice a sine in the limit, then this should immediately prompt us to think about the possibility of applying the first remarkable limit.

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

So, we have an uncertainty of the kind, its be sure to indicate in the design of the solution. The expression under the limit sign looks like the first remarkable limit, but this is not exactly it, it is located under the sine, but in the denominator.

In such cases, we need to organize the first remarkable limit ourselves, using an artificial trick. The line of reasoning can be as follows: “under the sine we have, it 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 the 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, the circled expression has turned into a unit and disappeared in the work:

Now it only remains to get rid of the three-story fraction:

If you have forgotten the simplification of multilevel fractions, please refresh the material in the reference book. Hot Formulas School Mathematics Course .

Ready. Final answer:

If you don't want to use pencil marks, then the solution can be made as follows:

“

Using the first wonderful limit

“

Example 2

Find the limit

Again we see a fraction and a sine in the limit. We try to substitute zero in the numerator and denominator:

Indeed, we have uncertainty and, therefore, we must try to organize the first remarkable limit. At the lesson Limits. Examples of solutions we considered the rule that when we have uncertainty, then we need to factor out the numerator and denominator. Here - the same thing, we will represent the degrees in the form of a product (factors):

Similarly to the previous example, we outline the remarkable limits with a pencil (there are two of them here), and indicate that they tend to unity:

Actually, the answer is ready:

In the following examples, I will not be engaged in arts in Paint, I think how to correctly draw up a solution in a notebook - you already understand.

Example 3

Find the limit

Substitute zero in the expression under the limit sign:

Uncertainty has been received that needs to be disclosed. If there is a tangent in the limit, then it is almost always converted into sine and cosine according to the well-known trigonometric formula (by the way, they do about the same with the cotangent, see. 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 is easy to get rid of it (do not forget to mark that it tends to one):

Thus, if in the limit the cosine is a MULTIPLIER, then, roughly speaking, it must be turned into a unit, which disappears in the product.

Here everything turned out easier, without any multiplication and division. The first remarkable limit also turns into one and disappears in the work:

As a result, infinity is obtained, it also happens.

Example 4

Find the limit

We try to substitute zero in 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, the limits with the use of this formula are very common.

We move the constant factors outside the limit icon:

Let's organize the first wonderful limit:

Here we have only one remarkable limit, which turns into a unit and disappears in the work:

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 wonderful limit by changing a variable, you can read about this a little later in the article Limit solving methods.

Second wonderful limit

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

This fact is called second wonderful limit.

Reference: Is an irrational number.

As a parameter, not only a variable can act, but also a complex function. It is only important that she strives for infinity..

Example 6

Find the limit

When the expression under the limit sign is in power, this is the first indication that a second remarkable limit should be attempted.

But first, as always, we try to substitute endlessly big number into an expression on what principle this is done, disassembled in the lesson Limits. Examples of solutions.

It is easy to see that for the base of the degree, and the exponent is , that is, there is an uncertainty of the form:

This uncertainty is just revealed with the help of the second remarkable limit. But, as often happens, the second remarkable limit does not lie on a silver platter, and must be artificially organized. You can argue as follows: in this example parameter, which means that we also need to organize in the indicator. To do this, we raise the base to a power, and so that the expression does not change, we raise it to a power:

When the task is completed by hand, we mark with a pencil:

Almost everything is ready, the terrible degree has turned into a pretty letter:

In this case, the limit icon itself is moved to the indicator:

Example 7

Find the limit

Attention! This type of limit is very common, please study this example very carefully.

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

The result is uncertainty. But a second remarkable limit applies to species uncertainty. What to do? You need to convert the base of the degree. We argue this way: in our denominator, it means that in the numerator we also need to organize.