home · On a note · The first remarkable limit. The second remarkable limit: examples of finding, problems and detailed solutions

The first remarkable limit. The second remarkable limit: examples of finding, problems and detailed solutions

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 an almost sure sign that this expression can be reduced to the 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 first remarkable limit is the following equality:

\begin(equation)\lim_(\alpha\to(0))\frac(\sin\alpha)(\alpha)=1 \end(equation)

Since for $\alpha\to(0)$ we have $\sin\alpha\to(0)$, they say that the first remarkable limit reveals an uncertainty of the form $\frac(0)(0)$. Generally speaking, in formula (1), instead of the variable $\alpha$, any expression can be placed under the sine sign and in the denominator, as long as two conditions are met:

  1. The expressions under the sine sign and in the denominator simultaneously tend to zero, i.e. there is uncertainty of the form $\frac(0)(0)$.
  2. The expressions under the sine sign and in the denominator are the same.

Corollaries from the first remarkable limit are also often used:

\begin(equation) \lim_(\alpha\to(0))\frac(\tg\alpha)(\alpha)=1 \end(equation) \begin(equation) \lim_(\alpha\to(0) )\frac(\arcsin\alpha)(\alpha)=1 \end(equation) \begin(equation) \lim_(\alpha\to(0))\frac(\arctg\alpha)(\alpha)=1 \end(equation)

Eleven examples are solved on this page. Example No. 1 is devoted to the proof of formulas (2)-(4). Examples No. 2, No. 3, No. 4 and No. 5 contain solutions with detailed comments. Examples No. 6-10 contain solutions with virtually no comments, because detailed explanations were given in previous examples. The solution uses some trigonometric formulas that can be found.

Let me note that the presence of trigonometric functions coupled with the uncertainty $\frac (0) (0)$ does not necessarily mean the application of the first remarkable limit. Sometimes simple trigonometric transformations are sufficient - for example, see.

Example No. 1

Prove that $\lim_(\alpha\to(0))\frac(\tg\alpha)(\alpha)=1$, $\lim_(\alpha\to(0))\frac(\arcsin\alpha )(\alpha)=1$, $\lim_(\alpha\to(0))\frac(\arctg\alpha)(\alpha)=1$.

a) Since $\tg\alpha=\frac(\sin\alpha)(\cos\alpha)$, then:

$$ \lim_(\alpha\to(0))\frac(\tg(\alpha))(\alpha)=\left|\frac(0)(0)\right| =\lim_(\alpha\to(0))\frac(\sin(\alpha))(\alpha\cos(\alpha)) $$

Since $\lim_(\alpha\to(0))\cos(0)=1$ and $\lim_(\alpha\to(0))\frac(\sin\alpha)(\alpha)=1$ , That:

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

b) Let's make the change $\alpha=\sin(y)$. Since $\sin(0)=0$, then from the condition $\alpha\to(0)$ we have $y\to(0)$. In addition, there is a neighborhood of zero in which $\arcsin\alpha=\arcsin(\sin(y))=y$, so:

$$ \lim_(\alpha\to(0))\frac(\arcsin\alpha)(\alpha)=\left|\frac(0)(0)\right| =\lim_(y\to(0))\frac(y)(\sin(y)) =\lim_(y\to(0))\frac(1)(\frac(\sin(y))( y)) =\frac(1)(\displaystyle\lim_(y\to(0))\frac(\sin(y))(y)) =\frac(1)(1) =1. $$

The equality $\lim_(\alpha\to(0))\frac(\arcsin\alpha)(\alpha)=1$ has been proven.

c) Let's make the replacement $\alpha=\tg(y)$. Since $\tg(0)=0$, then the conditions $\alpha\to(0)$ and $y\to(0)$ are equivalent. In addition, there is a neighborhood of zero in which $\arctg\alpha=\arctg\tg(y))=y$, therefore, based on the results of point a), we will have:

$$ \lim_(\alpha\to(0))\frac(\arctg\alpha)(\alpha)=\left|\frac(0)(0)\right| =\lim_(y\to(0))\frac(y)(\tg(y)) =\lim_(y\to(0))\frac(1)(\frac(\tg(y))( y)) =\frac(1)(\displaystyle\lim_(y\to(0))\frac(\tg(y))(y)) =\frac(1)(1) =1. $$

The equality $\lim_(\alpha\to(0))\frac(\arctg\alpha)(\alpha)=1$ has been proven.

Equalities a), b), c) are often used along with the first remarkable limit.

Example No. 2

Calculate the limit $\lim_(x\to(2))\frac(\sin\left(\frac(x^2-4)(x+7)\right))(\frac(x^2-4)( x+7))$.

Since $\lim_(x\to(2))\frac(x^2-4)(x+7)=\frac(2^2-4)(2+7)=0$ and $\lim_( x\to(2))\sin\left(\frac(x^2-4)(x+7)\right)=\sin(0)=0$, i.e. and both the numerator and denominator of the fraction simultaneously tend to zero, then here we are dealing with an uncertainty of the form $\frac(0)(0)$, i.e. done. In addition, it is clear that the expressions under the sine sign and in the denominator coincide (i.e., and is satisfied):

So, both conditions listed at the beginning of the page are met. It follows from this that the formula is applicable, i.e. $\lim_(x\to(2)) \frac(\sin\left(\frac(x^2-4)(x+7)\right))(\frac(x^2-4)(x+ 7))=1$.

Answer: $\lim_(x\to(2))\frac(\sin\left(\frac(x^2-4)(x+7)\right))(\frac(x^2-4)(x +7))=1$.

Example No. 3

Find $\lim_(x\to(0))\frac(\sin(9x))(x)$.

Since $\lim_(x\to(0))\sin(9x)=0$ and $\lim_(x\to(0))x=0$, then we are dealing with an uncertainty of the form $\frac(0 )(0)$, i.e. done. However, the expressions under the sine sign and in the denominator do not coincide. Here you need to adjust the expression in the denominator to the desired form. We need the expression $9x$ to be in the denominator, then it will become true. Essentially, we're missing a factor of $9$ in the denominator, which isn't that hard to enter—just multiply the expression in the denominator by $9$. Naturally, to compensate for multiplication by $9$, you will have to immediately divide by $9$:

$$ \lim_(x\to(0))\frac(\sin(9x))(x)=\left|\frac(0)(0)\right| =\lim_(x\to(0))\frac(\sin(9x))(9x\cdot\frac(1)(9)) =9\lim_(x\to(0))\frac(\sin (9x))(9x)$$

Now the expressions in the denominator and under the sine sign coincide. Both conditions for the limit $\lim_(x\to(0))\frac(\sin(9x))(9x)$ are satisfied. Therefore, $\lim_(x\to(0))\frac(\sin(9x))(9x)=1$. And this means that:

$$ 9\lim_(x\to(0))\frac(\sin(9x))(9x)=9\cdot(1)=9. $$

Answer: $\lim_(x\to(0))\frac(\sin(9x))(x)=9$.

Example No. 4

Find $\lim_(x\to(0))\frac(\sin(5x))(\tg(8x))$.

Since $\lim_(x\to(0))\sin(5x)=0$ and $\lim_(x\to(0))\tg(8x)=0$, here we are dealing with uncertainty of the form $\frac(0)(0)$. However, the form of the first remarkable limit is violated. A numerator containing $\sin(5x)$ requires a denominator of $5x$. In this situation, the easiest way is to divide the numerator by $5x$, and immediately multiply by $5x$. In addition, we will perform a similar operation with the denominator, multiplying and dividing $\tg(8x)$ by $8x$:

$$\lim_(x\to(0))\frac(\sin(5x))(\tg(8x))=\left|\frac(0)(0)\right| =\lim_(x\to(0))\frac(\frac(\sin(5x))(5x)\cdot(5x))(\frac(\tg(8x))(8x)\cdot(8x) )$$

Reducing by $x$ and taking the constant $\frac(5)(8)$ outside the limit sign, we get:

$$ \lim_(x\to(0))\frac(\frac(\sin(5x))(5x)\cdot(5x))(\frac(\tg(8x))(8x)\cdot(8x )) =\frac(5)(8)\cdot\lim_(x\to(0))\frac(\frac(\sin(5x))(5x))(\frac(\tg(8x))( 8x)) $$

Note that $\lim_(x\to(0))\frac(\sin(5x))(5x)$ fully satisfies the requirements for the first remarkable limit. To find $\lim_(x\to(0))\frac(\tg(8x))(8x)$ the following formula is applicable:

$$ \frac(5)(8)\cdot\lim_(x\to(0))\frac(\frac(\sin(5x))(5x))(\frac(\tg(8x))(8x )) =\frac(5)(8)\cdot\frac(\displaystyle\lim_(x\to(0))\frac(\sin(5x))(5x))(\displaystyle\lim_(x\to (0))\frac(\tg(8x))(8x)) =\frac(5)(8)\cdot\frac(1)(1) =\frac(5)(8). $$

Answer: $\lim_(x\to(0))\frac(\sin(5x))(\tg(8x))=\frac(5)(8)$.

Example No. 5

Find $\lim_(x\to(0))\frac(\cos(5x)-\cos^3(5x))(x^2)$.

Since $\lim_(x\to(0))(\cos(5x)-\cos^3(5x))=1-1=0$ (remember that $\cos(0)=1$) and $\lim_(x\to(0))x^2=0$, then we are dealing with uncertainty of the form $\frac(0)(0)$. However, in order to apply the first remarkable limit, you should get rid of the cosine in the numerator, moving on to sines (in order to then apply the formula) or tangents (in order to then apply the formula). This can be done with the following transformation:

$$\cos(5x)-\cos^3(5x)=\cos(5x)\cdot\left(1-\cos^2(5x)\right)$$ $$\cos(5x)-\cos ^3(5x)=\cos(5x)\cdot\left(1-\cos^2(5x)\right)=\cos(5x)\cdot\sin^2(5x).$$

Let's go back to the limit:

$$ \lim_(x\to(0))\frac(\cos(5x)-\cos^3(5x))(x^2)=\left|\frac(0)(0)\right| =\lim_(x\to(0))\frac(\cos(5x)\cdot\sin^2(5x))(x^2) =\lim_(x\to(0))\left(\cos (5x)\cdot\frac(\sin^2(5x))(x^2)\right) $$

The fraction $\frac(\sin^2(5x))(x^2)$ is already close to the form required for the first remarkable limit. Let's work a little with the fraction $\frac(\sin^2(5x))(x^2)$, adjusting it to the first remarkable limit (note that the expressions in the numerator and under the sine must match):

$$\frac(\sin^2(5x))(x^2)=\frac(\sin^2(5x))(25x^2\cdot\frac(1)(25))=25\cdot\ frac(\sin^2(5x))(25x^2)=25\cdot\left(\frac(\sin(5x))(5x)\right)^2$$

Let's return to the limit in question:

$$ \lim_(x\to(0))\left(\cos(5x)\cdot\frac(\sin^2(5x))(x^2)\right) =\lim_(x\to(0 ))\left(25\cos(5x)\cdot\left(\frac(\sin(5x))(5x)\right)^2\right)=\\ =25\cdot\lim_(x\to( 0))\cos(5x)\cdot\lim_(x\to(0))\left(\frac(\sin(5x))(5x)\right)^2 =25\cdot(1)\cdot( 1^2) =25. $$

Answer: $\lim_(x\to(0))\frac(\cos(5x)-\cos^3(5x))(x^2)=25$.

Example No. 6

Find the limit $\lim_(x\to(0))\frac(1-\cos(6x))(1-\cos(2x))$.

Since $\lim_(x\to(0))(1-\cos(6x))=0$ and $\lim_(x\to(0))(1-\cos(2x))=0$, then we are dealing with uncertainty $\frac(0)(0)$. Let us reveal it with the help of the first remarkable limit. To do this, let's move from cosines to sines. Since $1-\cos(2\alpha)=2\sin^2(\alpha)$, then:

$$1-\cos(6x)=2\sin^2(3x);\;1-\cos(2x)=2\sin^2(x).$$

Passing to sines in the given limit, we will have:

$$ \lim_(x\to(0))\frac(1-\cos(6x))(1-\cos(2x))=\left|\frac(0)(0)\right| =\lim_(x\to(0))\frac(2\sin^2(3x))(2\sin^2(x)) =\lim_(x\to(0))\frac(\sin^ 2(3x))(\sin^2(x))=\\ =\lim_(x\to(0))\frac(\frac(\sin^2(3x))((3x)^2)\ cdot(3x)^2)(\frac(\sin^2(x))(x^2)\cdot(x^2)) =\lim_(x\to(0))\frac(\left(\ frac(\sin(3x))(3x)\right)^2\cdot(9x^2))(\left(\frac(\sin(x))(x)\right)^2\cdot(x^ 2)) =9\cdot\frac(\displaystyle\lim_(x\to(0))\left(\frac(\sin(3x))(3x)\right)^2)(\displaystyle\lim_(x \to(0))\left(\frac(\sin(x))(x)\right)^2) =9\cdot\frac(1^2)(1^2) =9. $$

Answer: $\lim_(x\to(0))\frac(1-\cos(6x))(1-\cos(2x))=9$.

Example No. 7

Calculate the limit $\lim_(x\to(0))\frac(\cos(\alpha(x))-\cos(\beta(x)))(x^2)$ subject to $\alpha\neq\ beta$.

Detailed explanations were given earlier, but here we simply note that again there is uncertainty $\frac(0)(0)$. Let's move from cosines to sines using the formula

$$\cos\alpha-\cos\beta=-2\sin\frac(\alpha+\beta)(2)\cdot\sin\frac(\alpha-\beta)(2).$$

Using this formula, we get:

$$ \lim_(x\to(0))\frac(\cos(\alpha(x))-\cos(\beta(x)))(x^2)=\left|\frac(0)( 0)\right| =\lim_(x\to(0))\frac(-2\sin\frac(\alpha(x)+\beta(x))(2)\cdot\sin\frac(\alpha(x)-\ beta(x))(2))(x^2)=\\ =-2\cdot\lim_(x\to(0))\frac(\sin\left(x\cdot\frac(\alpha+\beta )(2)\right)\cdot\sin\left(x\cdot\frac(\alpha-\beta)(2)\right))(x^2) =-2\cdot\lim_(x\to( 0))\left(\frac(\sin\left(x\cdot\frac(\alpha+\beta)(2)\right))(x)\cdot\frac(\sin\left(x\cdot\frac (\alpha-\beta)(2)\right))(x)\right)=\\ =-2\cdot\lim_(x\to(0))\left(\frac(\sin\left(x \cdot\frac(\alpha+\beta)(2)\right))(x\cdot\frac(\alpha+\beta)(2))\cdot\frac(\alpha+\beta)(2)\cdot\frac (\sin\left(x\cdot\frac(\alpha-\beta)(2)\right))(x\cdot\frac(\alpha-\beta)(2))\cdot\frac(\alpha- \beta)(2)\right)=\\ =-\frac((\alpha+\beta)\cdot(\alpha-\beta))(2)\lim_(x\to(0))\frac(\ sin\left(x\cdot\frac(\alpha+\beta)(2)\right))(x\cdot\frac(\alpha+\beta)(2))\cdot\lim_(x\to(0)) \frac(\sin\left(x\cdot\frac(\alpha-\beta)(2)\right))(x\cdot\frac(\alpha-\beta)(2)) =-\frac(\ alpha^2-\beta^2)(2)\cdot(1)\cdot(1) =\frac(\beta^2-\alpha^2)(2). $$

Answer: $\lim_(x\to(0))\frac(\cos(\alpha(x))-\cos(\beta(x)))(x^2)=\frac(\beta^2-\ alpha^2)(2)$.

Example No. 8

Find the limit $\lim_(x\to(0))\frac(\tg(x)-\sin(x))(x^3)$.

Since $\lim_(x\to(0))(\tg(x)-\sin(x))=0$ (remember that $\sin(0)=\tg(0)=0$) and $\lim_(x\to(0))x^3=0$, then here we are dealing with uncertainty of the form $\frac(0)(0)$. Let's break it down as follows:

$$ \lim_(x\to(0))\frac(\tg(x)-\sin(x))(x^3)=\left|\frac(0)(0)\right| =\lim_(x\to(0))\frac(\frac(\sin(x))(\cos(x))-\sin(x))(x^3) =\lim_(x\to( 0))\frac(\sin(x)\cdot\left(\frac(1)(\cos(x))-1\right))(x^3) =\lim_(x\to(0)) \frac(\sin(x)\cdot\left(1-\cos(x)\right))(x^3\cdot\cos(x))=\\ =\lim_(x\to(0)) \frac(\sin(x)\cdot(2)\sin^2\frac(x)(2))(x^3\cdot\cos(x)) =\frac(1)(2)\cdot\ lim_(x\to(0))\left(\frac(\sin(x))(x)\cdot\left(\frac(\sin\frac(x)(2))(\frac(x)( 2))\right)^2\cdot\frac(1)(\cos(x))\right) =\frac(1)(2)\cdot(1)\cdot(1^2)\cdot(1 ) =\frac(1)(2). $$

Answer: $\lim_(x\to(0))\frac(\tg(x)-\sin(x))(x^3)=\frac(1)(2)$.

Example No. 9

Find the limit $\lim_(x\to(3))\frac(1-\cos(x-3))((x-3)\tg\frac(x-3)(2))$.

Since $\lim_(x\to(3))(1-\cos(x-3))=0$ and $\lim_(x\to(3))(x-3)\tg\frac(x -3)(2)=0$, then there is uncertainty of the form $\frac(0)(0)$. Before proceeding to its expansion, it is convenient to make a change of variable in such a way that the new variable tends to zero (note that in the formulas the variable $\alpha \to 0$). The easiest way is to introduce the variable $t=x-3$. However, for the sake of convenience of further transformations (this benefit can be seen in the course of the solution below), it is worth making the following replacement: $t=\frac(x-3)(2)$. I note that both replacements are applicable in this case, it’s just that the second replacement will allow you to work less with fractions. Since $x\to(3)$, then $t\to(0)$.

$$ \lim_(x\to(3))\frac(1-\cos(x-3))((x-3)\tg\frac(x-3)(2))=\left|\frac (0)(0)\right| =\left|\begin(aligned)&t=\frac(x-3)(2);\\&t\to(0)\end(aligned)\right| =\lim_(t\to(0))\frac(1-\cos(2t))(2t\cdot\tg(t)) =\lim_(t\to(0))\frac(2\sin^ 2t)(2t\cdot\tg(t)) =\lim_(t\to(0))\frac(\sin^2t)(t\cdot\tg(t))=\\ =\lim_(t\ to(0))\frac(\sin^2t)(t\cdot\frac(\sin(t))(\cos(t))) =\lim_(t\to(0))\frac(\sin (t)\cos(t))(t) =\lim_(t\to(0))\left(\frac(\sin(t))(t)\cdot\cos(t)\right) =\ lim_(t\to(0))\frac(\sin(t))(t)\cdot\lim_(t\to(0))\cos(t) =1\cdot(1) =1. $$

Answer: $\lim_(x\to(3))\frac(1-\cos(x-3))((x-3)\tg\frac(x-3)(2))=1$.

Example No. 10

Find the limit $\lim_(x\to\frac(\pi)(2))\frac(1-\sin(x))(\left(\frac(\pi)(2)-x\right)^2 )$.

Once again we are dealing with uncertainty $\frac(0)(0)$. Before proceeding to its expansion, it is convenient to make a change of variable in such a way that the new variable tends to zero (note that in the formulas the variable is $\alpha\to(0)$). The easiest way is to introduce the variable $t=\frac(\pi)(2)-x$. Since $x\to\frac(\pi)(2)$, then $t\to(0)$:

$$ \lim_(x\to\frac(\pi)(2))\frac(1-\sin(x))(\left(\frac(\pi)(2)-x\right)^2) =\left|\frac(0)(0)\right| =\left|\begin(aligned)&t=\frac(\pi)(2)-x;\\&t\to(0)\end(aligned)\right| =\lim_(t\to(0))\frac(1-\sin\left(\frac(\pi)(2)-t\right))(t^2) =\lim_(t\to(0 ))\frac(1-\cos(t))(t^2)=\\ =\lim_(t\to(0))\frac(2\sin^2\frac(t)(2))( t^2) =2\lim_(t\to(0))\frac(\sin^2\frac(t)(2))(t^2) =2\lim_(t\to(0))\ frac(\sin^2\frac(t)(2))(\frac(t^2)(4)\cdot(4)) =\frac(1)(2)\cdot\lim_(t\to( 0))\left(\frac(\sin\frac(t)(2))(\frac(t)(2))\right)^2 =\frac(1)(2)\cdot(1^2 ) =\frac(1)(2). $$

Answer: $\lim_(x\to\frac(\pi)(2))\frac(1-\sin(x))(\left(\frac(\pi)(2)-x\right)^2) =\frac(1)(2)$.

Example No. 11

Find the limits $\lim_(x\to\frac(\pi)(2))\frac(1-\sin(x))(\cos^2x)$, $\lim_(x\to\frac(2\ pi)(3))\frac(\tg(x)+\sqrt(3))(2\cos(x)+1)$.

In this case we don't have to use the first wonderful limit. Please note that both the first and second limits contain only trigonometric functions and numbers. Often in examples of this kind it is possible to simplify the expression located under the limit sign. Moreover, after the aforementioned simplification and reduction of some factors, the uncertainty disappears. I gave this example for only one purpose: to show that the presence of trigonometric functions under the limit sign does not necessarily mean the use of the first remarkable limit.

Since $\lim_(x\to\frac(\pi)(2))(1-\sin(x))=0$ (remember that $\sin\frac(\pi)(2)=1$ ) and $\lim_(x\to\frac(\pi)(2))\cos^2x=0$ (let me remind you that $\cos\frac(\pi)(2)=0$), then we have dealing with uncertainty of the form $\frac(0)(0)$. However, this does not mean that we will need to use the first wonderful limit. To reveal the uncertainty, it is enough to take into account that $\cos^2x=1-\sin^2x$:

$$ \lim_(x\to\frac(\pi)(2))\frac(1-\sin(x))(\cos^2x) =\left|\frac(0)(0)\right| =\lim_(x\to\frac(\pi)(2))\frac(1-\sin(x))(1-\sin^2x) =\lim_(x\to\frac(\pi)( 2))\frac(1-\sin(x))((1-\sin(x))(1+\sin(x))) =\lim_(x\to\frac(\pi)(2) )\frac(1)(1+\sin(x)) =\frac(1)(1+1) =\frac(1)(2). $$

There is a similar solution in Demidovich’s solution book (No. 475). As for the second limit, as in the previous examples in this section, we have an uncertainty of the form $\frac(0)(0)$. Why does it arise? It arises because $\tg\frac(2\pi)(3)=-\sqrt(3)$ and $2\cos\frac(2\pi)(3)=-1$. We use these values ​​to transform the expressions in the numerator and denominator. The goal of our actions is to write down the sum in the numerator and denominator as a product. By the way, often within a similar type it is convenient to change a variable, made in such a way that the new variable tends to zero (see, for example, examples No. 9 or No. 10 on this page). However, in this example there is no point in replacing, although if desired, replacing the variable $t=x-\frac(2\pi)(3)$ is not difficult to implement.

$$ \lim_(x\to\frac(2\pi)(3))\frac(\tg(x)+\sqrt(3))(2\cos(x)+1) =\lim_(x\ to\frac(2\pi)(3))\frac(\tg(x)+\sqrt(3))(2\cdot\left(\cos(x)+\frac(1)(2)\right )) =\lim_(x\to\frac(2\pi)(3))\frac(\tg(x)-\tg\frac(2\pi)(3))(2\cdot\left(\ cos(x)-\cos\frac(2\pi)(3)\right))=\\ =\lim_(x\to\frac(2\pi)(3))\frac(\frac(\sin \left(x-\frac(2\pi)(3)\right))(\cos(x)\cos\frac(2\pi)(3)))(-4\sin\frac(x+\frac (2\pi)(3))(2)\sin\frac(x-\frac(2\pi)(3))(2)) =\lim_(x\to\frac(2\pi)(3 ))\frac(\sin\left(x-\frac(2\pi)(3)\right))(-4\sin\frac(x+\frac(2\pi)(3))(2)\ sin\frac(x-\frac(2\pi)(3))(2)\cos(x)\cos\frac(2\pi)(3))=\\ =\lim_(x\to\frac (2\pi)(3))\frac(2\sin\frac(x-\frac(2\pi)(3))(2)\cos\frac(x-\frac(2\pi)(3 ))(2))(-4\sin\frac(x+\frac(2\pi)(3))(2)\sin\frac(x-\frac(2\pi)(3))(2) \cos(x)\cos\frac(2\pi)(3)) =\lim_(x\to\frac(2\pi)(3))\frac(\cos\frac(x-\frac(2 \pi)(3))(2))(-2\sin\frac(x+\frac(2\pi)(3))(2)\cos(x)\cos\frac(2\pi)(3 ))=\\ =\frac(1)(-2\cdot\frac(\sqrt(3))(2)\cdot\left(-\frac(1)(2)\right)\cdot\left( -\frac(1)(2)\right)) =-\frac(4)(\sqrt(3)). $$

As you can see, we didn't have to apply the first wonderful limit. Of course, you can do this if you want (see note below), but it is not necessary.

What is the solution using the first remarkable limit? show\hide

Using the first remarkable limit we get:

$$ \lim_(x\to\frac(2\pi)(3))\frac(\sin\left(x-\frac(2\pi)(3)\right))(-4\sin\frac (x+\frac(2\pi)(3))(2)\sin\frac(x-\frac(2\pi)(3))(2)\cos(x)\cos\frac(2\pi )(3))=\\ =\lim_(x\to\frac(2\pi)(3))\left(\frac(\sin\left(x-\frac(2\pi)(3)\ right))(x-\frac(2\pi)(3))\cdot\frac(1)(\frac(\sin\frac(x-\frac(2\pi)(3))(2)) (\frac(x-\frac(2\pi)(3))(2)))\cdot\frac(1)(-2\sin\frac(x+\frac(2\pi)(3))( 2)\cos(x)\cos\frac(2\pi)(3))\right) =1\cdot(1)\cdot\frac(1)(-2\cdot\frac(\sqrt(3) )(2)\cdot\left(-\frac(1)(2)\right)\cdot\left(-\frac(1)(2)\right)) =-\frac(4)(\sqrt( 3)). $$

Answer: $\lim_(x\to\frac(\pi)(2))\frac(1-\sin(x))(\cos^2x)=\frac(1)(2)$, $\lim_( x\to\frac(2\pi)(3))\frac(\tg(x)+\sqrt(3))(2\cos(x)+1)=-\frac(4)(\sqrt( 3))$.

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 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 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 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!

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 remarkable 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, we substitute zero instead of 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 what 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.
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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) 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 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 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