Remarkable limits. Examples of solutions. Second wonderful limit

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. In the course of mathematical analysis, it is proven that:

This mathematical fact is called The first wonderful limit. I won’t give an analytical proof of the limit, but we’ll look at its geometric meaning in the lesson 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, but also an elementary function or a complex function can act as a parameter. 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 a “two” and a “three”, and the teacher decides to ask the student some simple question or offer to solve a simple 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 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 an infinitely large number into the expression, the principle by which this is done is 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. You can reason as follows: in this example the parameter is , which means that we also need to organize in the indicator. 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 .

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 the general formulas of remarkable limits themselves 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 of the first remarkable limit, and then 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 that is 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 have difficulty with limits, it will help them better understand the rules and master the 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 limit of the 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. The trigonometric formula should be used here

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, which is 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 obtain 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, deriving 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!

Proof:

Let us first prove the theorem for the case of the sequence

According to Newton's binomial formula:

Assuming we get

From this equality (1) it follows that as n increases, the number of positive terms on the right side increases. In addition, as n increases, the number decreases, so the values are increasing. Therefore the sequence increasing, and (2)*We show that it is bounded. Replace each parenthesis on the right side of the equality with one, the right side will increase, and we get the inequality

Let's strengthen the resulting inequality, replace 3,4,5, ..., standing in the denominators of the fractions, with the number 2: We find the sum in brackets using the formula for the sum of the terms of a geometric progression: Therefore (3)*

So, the sequence is bounded from above, and inequalities (2) and (3) are satisfied: Therefore, based on the Weierstrass theorem (criterion for the convergence of a sequence), the sequence monotonically increases and is 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 prove that . Let's consider two cases:

1. Let Each value of x be enclosed between two positive integers: , where is the integer part of x. => =>

If , then Therefore, according to the limit We have

Based on the criterion (about the limit of an intermediate function) of the existence of limits

2. Let . Let's make the substitution − x = t, then

From these two cases it follows that for real x.

Consequences:

9 .) Comparison of infinitesimals. The theorem on replacing infinitesimals with equivalent ones in the limit and the theorem on the main part of infinitesimals.

Let the functions a( x) and b( x) – b.m. at x ® x 0 .

DEFINITIONS.

1)a( x) called infinitesimal higher order than b (x) If

Write down: a( x) = o(b( x)) .

2)a( x) And b( x)are called infinitesimals of the same order, If

where CÎℝ and C¹ 0 .

Write down: a( x) = O(b( x)) .

3)a( x) And b( x) are called equivalent , If

Write down: a( x) ~ b( x).

4)a( x) is called infinitesimal of order k relative
absolutely infinitesimal b( x), if infinitesimal a( x)And(b( x))k have the same order, i.e. If

where CÎℝ and C¹ 0 .

THEOREM 6 (on replacing infinitesimals with equivalent ones).

Let 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),

That

Proof: Let a( x) ~ a 1 ( x), b( x) ~ b 1 ( x), Then

THEOREM 7 (about the main part of the infinitesimal).

Let 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) – 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, geometric limits) One-sided continuity. Continuity on an interval, on a segment. Properties of continuous functions.

1. Basic definitions

Let f(x) is defined in some neighborhood of the point x 0 .

DEFINITION 1. Function f(x) called continuous at a point x 0 if the equality is true

Notes.

1) By virtue of Theorem 5 §3, equality (1) can be written in the form

Condition (2) – definition of 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 a function is continuous at a point x 0, then the sign of the limit and the function can be swapped."

DEFINITION 2 (in e-d language).

Function f(x) called continuous at a point x 0 If"e>0 $d>0 such, What

if xОU( 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 x, x 0 Î D(f) (x 0 – fixed, x – arbitrary)

Let's denote: D x= x – x 0 – argument increment

D f(x 0) = f(x) – f(x 0) – increment of function at pointx 0

DEFINITION 3 (geometric).

Function f(x) on called continuous at a point x 0 if at this point an infinitesimal increment in the argument corresponds to an infinitesimal increment in 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 a 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-way continuity at boundary points(i.e. continuous at the point a on the right, at the point b- left).

11) Break points, their classification

DEFINITION. If function f(x) defined in some neighborhood of point x 0 , but is not continuous at this point, then f(x) called discontinuous at point x 0 , and the point itself x 0 called the break point functions f(x) .

Notes.

1) f(x) can be defined in an incomplete neighborhood of the point x 0 .

Then consider the corresponding one-sided continuity of the function.

2) From the definition of Þ point x 0 is the break point of the function f(x) in two cases:

a) U( x 0 , d)О D(f) , but for f(x) equality does not hold

b) U * ( x 0 , d)О D(f) .

For elementary functions, only case b) is possible.

Let x 0 – function break point f(x) .

DEFINITION. Point x 0 called break point I sort of if function f(x)has finite limits on the left and right at this point.

If these limits are equal, then point x 0 called removable break point , otherwise - jump point .

DEFINITION. Point x 0 called break point II sort of if at least one of the one-sided limits of the function f(x)at this point is equal¥ or doesn't exist.

12) Properties of functions continuous on an interval (theorems of Weierstrass (without proof) and Cauchy

Weierstrass's theorem

Let the function f(x) be continuous on the interval, then

1)f(x)is limited to

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 said to be greatest if m≥f(x) for any x € D(f).

The function can take on 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 the segment and x be the number contained between f(a) and f(b), then there is at least one point x 0 € such that f(x 0)= g

Now, with a calm soul, let’s move on to consider wonderful limits.
looks like .

Instead of the variable x, various functions can be present, the main thing is that they tend to 0.

It is necessary to calculate the limit

As you can see, this limit is very similar to the first wonderful one, but this is not entirely true. In general, if you notice sin in the limit, then you should immediately think about whether it is possible to use the first remarkable limit.

According to our rule No. 1, let’s substitute zero for x:

We get uncertainty.

Now let's try to organize the first wonderful limit ourselves. To do this, let's do a simple combination:

So we organize the numerator and denominator to highlight 7x. Now the familiar remarkable limit has already appeared. It is advisable to highlight it when deciding:

Let's substitute the solution to the first remarkable example and get:

Simplifying the fraction:

Answer: 7/3.

As you can see, everything is very simple.

Looks like , where e = 2.718281828... is an irrational number.

Various functions may be present instead of the variable x, the main thing is that they tend to .

It is necessary to calculate the limit

Here we see the presence of a degree under the sign of a limit, which means it is possible to use a second remarkable limit.

As always, we will use rule No. 1 - substitute x instead of:

It can be seen that at x the base of the degree is , and the exponent is 4x > , i.e. we obtain an uncertainty of the form:

Let's use the second wonderful limit to reveal our uncertainty, but first we need to organize it. As you can see, we need to achieve presence in the indicator, for which we raise the base to the power of 3x, and at the same time to the power of 1/3x, so that the expression does not change:

Don't forget to highlight our wonderful limit:

That's how they really are wonderful limits!
If you still have any questions about the first and second wonderful limits, then feel free to ask them in the comments.
We will answer everyone as much as possible.

You can also work with a teacher on this topic.
We are pleased to offer you the services of selecting a qualified tutor in your city. Our partners will quickly select a good teacher for you on favorable terms.

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The first remarkable limit is often used to calculate limits containing sine, arcsine, tangent, arctangent and the resulting uncertainties of zero divided by zero.

Formula

The formula for the first remarkable limit is: $$ \lim_(\alpha\to 0) \frac(\sin\alpha)(\alpha) = 1 $$

We note that for $ \alpha\to 0 $ we get $ \sin\alpha \to 0 $, thus we have zeros in the numerator and denominator. Thus, the formula of the first remarkable limit is needed to reveal the uncertainties $ \frac(0)(0) $.

To apply the formula, two conditions must be met:

1. The expressions contained in the sine and the denominator of the fraction are the same
2. Expressions in the sine and denominator of a fraction tend to zero

Attention! $ \lim_(x\to 0) \frac(\sin(2x^2+1))(2x^2+1) \neq 1 $ Although the expressions under the sine and in the denominator are the same, however $ 2x^2+1 = 1 $, for $ x\to 0 $. The second condition is not met, so you CANNOT apply the formula!

Consequences

Quite rarely in tasks you can see a pure first wonderful limit, in which you could immediately write down the answer. In practice, everything looks a little more complicated, but for such cases it will be useful to know the consequences of the first remarkable limit. Thanks to them, you can quickly calculate the required limits.

$$ \lim_(\alpha\to 0) \frac(\alpha)(\sin\alpha) = 1 $$

$$ \lim_(\alpha\to 0) \frac(\sin(a\alpha))(\sin(b\alpha)) = \frac(a)(b) $$

$$ \lim_(\alpha\to 0) \frac(tg\alpha)(\alpha) = 1 $$

$$ \lim_(\alpha\to 0) \frac(\arcsin\alpha)(\alpha) = 1 $$

$$ \lim_(\alpha\to 0) \frac(arctg\alpha)(\alpha) = 1 $$

Examples of solutions

Let's consider the first remarkable limit, examples of its solution for calculating limits containing trigonometric functions and uncertainty $ \bigg[\frac(0)(0)\bigg] $

Example 1

Calculate $ \lim_(x\to 0) \frac(\sin2x)(4x) $

Solution

Let's look at the limit and notice that it contains a sine. Next, we substitute $ x = 0 $ into the numerator and denominator and get the uncertainty zero divided by zero: $$ \lim_(x\to 0) \frac(\sin2x)(4x) = \frac(0)(0) $$ Already two signs that we need to apply a wonderful limit, but there is a small nuance: we cannot immediately apply the formula, since the expression under the sine sign differs from the expression in the denominator. And we need them to be equal. Therefore, with the help of elementary transformations of the numerator we will turn it into $2x$. To do this, we will take the two out of the denominator of the fraction as a separate factor. It looks like this: $$ \lim_(x\to 0) \frac(\sin2x)(4x) = \lim_(x\to 0) \frac(\sin2x)(2\cdot 2x) = $$ $$ = \frac(1)(2) \lim_(x\to 0) \frac(\sin2x)(2x) = \frac(1)(2)\cdot 1 = \frac(1)(2) $$ Please note , that at the end $ \lim_(x\to 0) \frac(\sin2x)(2x) = 1 $ was obtained according to the formula. If you cannot solve your problem, then send it to us. We will provide detailed solution. You will be able to view the progress of the calculation and gain information. This will help you get your grade from your teacher in a timely manner!

Answer

$$ \lim_(x\to 0) \frac(\sin2x)(4x) =\frac(1)(2) $$

Example 2

Find $ \lim_(x\to 0) \frac(\sin(x^3+2x))(2x-x^4) $

Solution

As always, you first need to know the type of uncertainty. If it is zero divided by zero, then we pay attention to the presence of a sine: $$ \lim_(x\to 0) \frac(\sin(x^3+2x))(2x-x^4) = \frac(0) (0) = $$ This uncertainty allows us to use the formula of the first remarkable limit, but the expression from the denominator is not equal to the argument of the sine? Therefore, the formula cannot be applied “head-on”. It is necessary to multiply and divide the fraction by the argument of the sine: $$ = \lim_(x\to 0) \frac((x^3+2x)\sin(x^3+2x))((2x-x^4)(x ^3+2x)) = $$ Now we write down the properties of the limits: $$ = \lim_(x\to 0) \frac((x^3+2x))(2x-x^4)\cdot \lim_(x \to 0) \frac(\sin(x^3+2x))((x^3+2x)) = $$ The second limit fits the formula exactly and is equal to one: $$ = \lim_(x\to 0 ) \frac(x^3+2x)(2x-x^4)\cdot 1 = \lim_(x\to 0) \frac(x^3+2x)(2x-x^4) = $$ Substitute again $ x = 0 $ into a fraction and we get the uncertainty $ \frac(0)(0) $. To eliminate it, it is enough to take $ x $ out of brackets and reduce it by: $$ = \lim_(x\to 0) \frac(x(x^2+2))(x(2-x^3)) = \ lim_(x\to 0) \frac(x^2+2)(2-x^3) = $$ $$ = \frac(0^2 + 2)(2 - 0^3) = \frac(2 )(2) = 1 $$

Answer

$$ \lim_(x\to 0) \frac(\sin(x^3+2x))(2x-x^4) = 1 $$

Example 4

Calculate $ \lim_(x\to0) \frac(\sin2x)(tg3x) $

Solution

Let's start the calculation with the substitution $ x=0 $. As a result, we obtain the uncertainty $ \frac(0)(0) $. The limit contains a sine and a tangent, which hints at a possible development of the situation using the formula of the first remarkable limit. Let's transform the numerator and denominator of the fraction into a formula and consequence: $$ \lim_(x\to0) \frac(\sin2x)(tg3x) = \frac(0)(0) = \lim_(x\to0) \frac(\frac(\sin2x)(2x)\cdot 2x )(\frac(tg3x)(3x)\cdot 3x) = $$ Now we see that in the numerator and denominator there are expressions that fit the formula and consequences. The sine argument and the tangent argument are the same for the corresponding denominators $$ = \lim_(x\to0) \frac(1\cdot 2x)(1\cdot 3x) = \frac(2)(3) $$

Answer

$$ \lim_(x\to0) \frac(\sin2x)(tg2x) = \frac(2)(3) $$

The article: “The first remarkable limit, examples of solutions” talked about cases in which it is advisable to use this formula and its consequences.