The second remarkable limit. First wonderful limit 1 wonderful limit calculator

The formula for the second remarkable limit is lim x → ∞ 1 + 1 x x = e. Another form of writing looks like this: lim x → 0 (1 + x) 1 x = e.

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

Let's consider problems in which the ability to calculate the second remarkable limit will be useful.

Example 1

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

Solution

Let's substitute the required formula and perform the calculations.

lim x → ∞ 1 - 2 x 2 + 1 x 2 + 1 4 = 1 - 2 ∞ 2 + 1 ∞ 2 + 1 4 = 1 - 0 ∞ = 1 ∞

Our answer turned out to be one to the power of infinity. To determine the solution method, we use the uncertainty table. Let's choose the second remarkable limit and make a change of variables.

t = - x 2 + 1 2 ⇔ x 2 + 1 4 = - t 2

If x → ∞, then t → - ∞.

Let's see what we got after the replacement:

lim x → ∞ 1 - 2 x 2 + 1 x 2 + 1 4 = 1 ∞ = lim x → ∞ 1 + 1 t - 1 2 t = lim t → ∞ 1 + 1 t t - 1 2 = e - 1 2

Answer: lim x → ∞ 1 - 2 x 2 + 1 x 2 + 1 4 = e - 1 2 .

Example 2

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

Solution

Let's substitute infinity and get the following.

lim x → ∞ x - 1 x + 1 x = lim x → ∞ 1 - 1 x 1 + 1 x x = 1 - 0 1 + 0 ∞ = 1 ∞

In the answer, we again got the same thing as in the previous problem, therefore, we can again use the second remarkable limit. Next, we need to select the whole part at the base of the power function:

x - 1 x + 1 = x + 1 - 2 x + 1 = x + 1 x + 1 - 2 x + 1 = 1 - 2 x + 1

After this, the limit takes the following form:

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

Replace variables. Let's assume that t = - x + 1 2 ⇒ 2 t = - x - 1 ⇒ x = - 2 t - 1 ; if x → ∞, then t → ∞.

After that, we write down what we got in the original limit:

lim x → ∞ x - 1 x + 1 x = 1 ∞ = lim x → ∞ 1 - 2 x + 1 x = lim x → ∞ 1 + 1 t - 2 t - 1 = = lim x → ∞ 1 + 1 t - 2 t 1 + 1 t - 1 = lim x → ∞ 1 + 1 t - 2 t lim x → ∞ 1 + 1 t - 1 = = lim x → ∞ 1 + 1 t t - 2 1 + 1 ∞ = e - 2 · (1 + 0) - 1 = e - 2

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

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

Example 3

Calculate the limit lim x → ∞ x 3 + 1 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 .

Solution

lim x → ∞ x 3 + 1 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 = lim x → ∞ 1 + 1 x 3 1 + 2 x - 1 x 3 3 2 x - 5 x 4 = = 1 + 0 1 + 0 - 0 3 0 - 0 = 1 ∞

After that, we need to transform the function to apply the second great limit. We got the following:

lim x → ∞ x 3 + 1 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 = 1 ∞ = lim x → ∞ x 3 - 2 x 2 - 1 - 2 x 2 + 2 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 = = lim x → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5

lim x → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 = lim x → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 x 3 + 2 x 2 - 1 - 2 x 2 + 2 - 2 x 2 + 2 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 = = lim x → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 x 3 + 2 x 2 - 1 - 2 x 2 + 2 - 2 x 2 + 2 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5

Since we now have the same exponents in the numerator and denominator of the fraction (equal to six), the limit of the fraction at infinity will be equal to the ratio of these coefficients at higher powers.

lim x → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 x 3 + 2 x 2 - 1 - 2 x 2 + 2 - 2 x 2 + 2 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 = = lim x → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 x 3 + 2 x 2 - 1 - 2 x 2 + 2 - 6 2 = lim x → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 x 3 + 2 x 2 - 1 - 2 x 2 + 2 - 3

By substituting t = x 2 + 2 x 2 - 1 - 2 x 2 + 2 we get a second remarkable limit. Means what:

lim x → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 x 3 + 2 x 2 - 1 - 2 x 2 + 2 - 3 = lim x → ∞ 1 + 1 t t - 3 = e - 3

Answer: lim x → ∞ x 3 + 1 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 = e - 3 .

conclusions

Uncertainty 1 ∞, i.e. unity to an infinite power is a power-law uncertainty, therefore, it can be revealed using the rules for finding the limits of exponential power functions.

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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] $

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

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.

Usually the second remarkable limit is written in this form:

\begin(equation) \lim_(x\to\infty)\left(1+\frac(1)(x)\right)^x=e\end(equation)

The number $e$ indicated on the right side of equality (1) is irrational. The approximate value of this number is: $e\approx(2(,)718281828459045)$. If we make the replacement $t=\frac(1)(x)$, then formula (1) can be rewritten as follows:

\begin(equation) \lim_(t\to(0))\biggl(1+t\biggr)^(\frac(1)(t))=e\end(equation)

As for the first remarkable limit, it does not matter which expression stands in place of the variable $x$ in formula (1) or instead of the variable $t$ in formula (2). The main thing is to fulfill two conditions:

1. The base of the degree (i.e., the expression in brackets of formulas (1) and (2)) should tend to unity;
2. The exponent (i.e. $x$ in formula (1) or $\frac(1)(t)$ in formula (2)) must tend to infinity.

The second remarkable limit is said to reveal the uncertainty of $1^\infty$. Please note that in formula (1) we do not specify which infinity ($+\infty$ or $-\infty$) we are talking about. In any of these cases, formula (1) is correct. In formula (2), the variable $t$ can tend to zero both on the left and on the right.

I note that there are also several useful consequences from the second remarkable limit. Examples of the use of the second remarkable limit, as well as its consequences, are very popular among compilers of standard standard calculations and tests.

Example No. 1

Calculate the limit $\lim_(x\to\infty)\left(\frac(3x+1)(3x-5)\right)^(4x+7)$.

Let us immediately note that the base of the degree (i.e. $\frac(3x+1)(3x-5)$) tends to unity:

$$ \lim_(x\to\infty)\frac(3x+1)(3x-5)=\left|\frac(\infty)(\infty)\right| =\lim_(x\to\infty)\frac(3+\frac(1)(x))(3-\frac(5)(x)) =\frac(3+0)(3-0) = 1. $$

In this case, the exponent (expression $4x+7$) tends to infinity, i.e. $\lim_(x\to\infty)(4x+7)=\infty$.

The base of the degree tends to unity, the exponent tends to infinity, i.e. we are dealing with uncertainty $1^\infty$. Let's apply a formula to reveal this uncertainty. At the base of the power of the formula is the expression $1+\frac(1)(x)$, and in the example we are considering, the base of the power is: $\frac(3x+1)(3x-5)$. Therefore, the first action will be a formal adjustment of the expression $\frac(3x+1)(3x-5)$ to the form $1+\frac(1)(x)$. First, add and subtract one:

$$ \lim_(x\to\infty)\left(\frac(3x+1)(3x-5)\right)^(4x+7) =|1^\infty| =\lim_(x\to\infty)\left(1+\frac(3x+1)(3x-5)-1\right)^(4x+7) $$

Please note that you cannot simply add a unit. If we are forced to add one, then we also need to subtract it so as not to change the value of the entire expression. To continue the solution, we take into account that

$$ \frac(3x+1)(3x-5)-1 =\frac(3x+1)(3x-5)-\frac(3x-5)(3x-5) =\frac(3x+1- 3x+5)(3x-5) =\frac(6)(3x-5). $$

Since $\frac(3x+1)(3x-5)-1=\frac(6)(3x-5)$, then:

$$ \lim_(x\to\infty)\left(1+ \frac(3x+1)(3x-5)-1\right)^(4x+7) =\lim_(x\to\infty)\ left(1+\frac(6)(3x-5)\right)^(4x+7) $$

Let's continue the adjustment. In the expression $1+\frac(1)(x)$ of the formula, the numerator of the fraction is 1, and in our expression $1+\frac(6)(3x-5)$ the numerator is $6$. To get $1$ in the numerator, drop $6$ into the denominator using the following conversion:

$$ 1+\frac(6)(3x-5) =1+\frac(1)(\frac(3x-5)(6)) $$

Thus,

$$ \lim_(x\to\infty)\left(1+\frac(6)(3x-5)\right)^(4x+7) =\lim_(x\to\infty)\left(1+ \frac(1)(\frac(3x-5)(6))\right)^(4x+7) $$

So, the basis of the degree, i.e. $1+\frac(1)(\frac(3x-5)(6))$, adjusted to the form $1+\frac(1)(x)$ required in the formula. Now let's start working with the exponent. Note that in the formula the expressions in the exponents and in the denominator are the same:

This means that in our example, the exponent and the denominator must be brought to the same form. To get the expression $\frac(3x-5)(6)$ in the exponent, we simply multiply the exponent by this fraction. Naturally, to compensate for such a multiplication, you will have to immediately multiply by the reciprocal fraction, i.e. by $\frac(6)(3x-5)$. So we have:

$$ \lim_(x\to\infty)\left(1+\frac(1)(\frac(3x-5)(6))\right)^(4x+7) =\lim_(x\to\ infty)\left(1+\frac(1)(\frac(3x-5)(6))\right)^(\frac(3x-5)(6)\cdot\frac(6)(3x-5 )\cdot(4x+7)) =\lim_(x\to\infty)\left(\left(1+\frac(1)(\frac(3x-5)(6))\right)^(\ frac(3x-5)(6))\right)^(\frac(6\cdot(4x+7))(3x-5)) $$

Let us separately consider the limit of the fraction $\frac(6\cdot(4x+7))(3x-5)$ located in the power:

$$ \lim_(x\to\infty)\frac(6\cdot(4x+7))(3x-5) =\left|\frac(\infty)(\infty)\right| =\lim_(x\to\infty)\frac(6\cdot\left(4+\frac(7)(x)\right))(3-\frac(5)(x)) =6\cdot\ frac(4)(3) =8. $$

Answer: $\lim_(x\to(0))\biggl(\cos(2x)\biggr)^(\frac(1)(\sin^2(3x)))=e^(-\frac(2) (9))$.

Example No. 4

Find the limit $\lim_(x\to+\infty)x\left(\ln(x+1)-\ln(x)\right)$.

Since for $x>0$ we have $\ln(x+1)-\ln(x)=\ln\left(\frac(x+1)(x)\right)$, then:

$$ \lim_(x\to+\infty)x\left(\ln(x+1)-\ln(x)\right) =\lim_(x\to+\infty)\left(x\cdot\ln\ left(\frac(x+1)(x)\right)\right) $$

Expanding the fraction $\frac(x+1)(x)$ into the sum of fractions $\frac(x+1)(x)=1+\frac(1)(x)$ we get:

$$ \lim_(x\to+\infty)\left(x\cdot\ln\left(\frac(x+1)(x)\right)\right) =\lim_(x\to+\infty)\left (x\cdot\ln\left(1+\frac(1)(x)\right)\right) =\lim_(x\to+\infty)\left(\ln\left(\frac(x+1) (x)\right)^x\right) =\ln(e) =1. $$

Answer: $\lim_(x\to+\infty)x\left(\ln(x+1)-\ln(x)\right)=1$.

Example No. 5

Find the limit $\lim_(x\to(2))\biggl(3x-5\biggr)^(\frac(2x)(x^2-4))$.

Since $\lim_(x\to(2))(3x-5)=6-5=1$ and $\lim_(x\to(2))\frac(2x)(x^2-4)= \infty$, then we are dealing with uncertainty of the form $1^\infty$. Detailed explanations are given in example No. 2, but here we will limit ourselves to a brief solution. Making the replacement $t=x-2$, we get:

$$ \lim_(x\to(2))\biggl(3x-5\biggr)^(\frac(2x)(x^2-4)) =\left|\begin(aligned)&t=x-2 ;\;x=t+2\\&t\to(0)\end(aligned)\right| =\lim_(t\to(0))\biggl(1+3t\biggr)^(\frac(2t+4)(t^2+4t))=\\ =\lim_(t\to(0) )\biggl(1+3t\biggr)^(\frac(1)(3t)\cdot 3t\cdot\frac(2t+4)(t^2+4t)) =\lim_(t\to(0) )\left(\biggl(1+3t\biggr)^(\frac(1)(3t))\right)^(\frac(6\cdot(t+2))(t+4)) =e^ 3. $$

You can solve this example in a different way, using the replacement: $t=\frac(1)(x-2)$. Of course, the answer will be the same:

$$ \lim_(x\to(2))\biggl(3x-5\biggr)^(\frac(2x)(x^2-4)) =\left|\begin(aligned)&t=\frac( 1)(x-2);\;x=\frac(2t+1)(t)\\&t\to\infty\end(aligned)\right| =\lim_(t\to\infty)\left(1+\frac(3)(t)\right)^(t\cdot\frac(4t+2)(4t+1))=\\ =\lim_ (t\to\infty)\left(1+\frac(1)(\frac(t)(3))\right)^(\frac(t)(3)\cdot\frac(3)(t) \cdot\frac(t\cdot(4t+2))(4t+1)) =\lim_(t\to\infty)\left(\left(1+\frac(1)(\frac(t)( 3))\right)^(\frac(t)(3))\right)^(\frac(6\cdot(2t+1))(4t+1)) =e^3. $$

Answer: $\lim_(x\to(2))\biggl(3x-5\biggr)^(\frac(2x)(x^2-4))=e^3$.

Example No. 6

Find the limit $\lim_(x\to\infty)\left(\frac(2x^2+3)(2x^2-4)\right)^(3x) $.

Let's find out what the expression $\frac(2x^2+3)(2x^2-4)$ tends to under the condition $x\to\infty$:

$$ \lim_(x\to\infty)\frac(2x^2+3)(2x^2-4) =\left|\frac(\infty)(\infty)\right| =\lim_(x\to\infty)\frac(2+\frac(3)(x^2))(2-\frac(4)(x^2)) =\frac(2+0)(2 -0)=1. $$

Thus, in the given limit we are dealing with an uncertainty of the form $1^\infty$, which we will reveal using the second remarkable limit:

$$ \lim_(x\to\infty)\left(\frac(2x^2+3)(2x^2-4)\right)^(3x) =|1^\infty| =\lim_(x\to\infty)\left(1+\frac(2x^2+3)(2x^2-4)-1\right)^(3x)=\\ =\lim_(x\to \infty)\left(1+\frac(7)(2x^2-4)\right)^(3x) =\lim_(x\to\infty)\left(1+\frac(1)(\frac (2x^2-4)(7))\right)^(3x)=\\ =\lim_(x\to\infty)\left(1+\frac(1)(\frac(2x^2-4 )(7))\right)^(\frac(2x^2-4)(7)\cdot\frac(7)(2x^2-4)\cdot 3x) =\lim_(x\to\infty) \left(\left(1+\frac(1)(\frac(2x^2-4)(7))\right)^(\frac(2x^2-4)(7))\right)^( \frac(21x)(2x^2-4)) =e^0 =1. $$

Answer: $\lim_(x\to\infty)\left(\frac(2x^2+3)(2x^2-4)\right)^(3x)=1$.

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