The function limit is wonderful limits. Second wonderful limit
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:
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The denominator is sine, therefore, the expression can be brought to the first remarkable limit. Let's start the transformation:
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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 wonderful limit:
because it doesn’t matter which letter (variable) in this formula stands instead of X.
We multiply X by three and immediately divide:
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In accordance with the first remarkable limit noticed, we replace the fractional expression:
Now we can finally solve this limit:
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Example 2. Find the limit.
Solution. Direct substitution again leads to the “zero divided by zero” uncertainty:
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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:
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Example 3. Find the limit.
Solution. When substituting, we again get the uncertainty “zero divided by zero”:
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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:
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Example 4. Find the limit.
Solution. Once again we get the uncertainty “zero divided by zero”:
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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:
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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.
In this topic, we will analyze the formulas that can be obtained using the second remarkable limit (a topic dedicated directly to the second remarkable limit is located). Let me recall two formulations of the second remarkable limit that will be needed in this section: $\lim_(x\to\infty)\left(1+\frac(1)(x)\right)^x=e$ and $\lim_(x \to\ 0)\left(1+x\right)^\frac(1)(x)=e$.
Usually I present formulas without proof, but for this page, I think I’ll make an exception. The point is that the proof of the consequences of the second remarkable limit contains some techniques that are useful in directly solving problems. Well, generally speaking, it is advisable to know how this or that formula is proven. This allows you to better understand its internal structure, as well as the limits of applicability. But since the evidence may not be of interest to all readers, I will hide it under the notes located after each consequence.
Corollary #1
\begin(equation) \lim_(x\to\ 0) \frac(\ln(1+x))(x)=1\end(equation)Evidence of corollary No. 1: show\hide
Since at $x\to 0$ we have $\ln(1+x)\to 0$, then in the limit under consideration there is an uncertainty of the form $\frac(0)(0)$. To reveal this uncertainty, let us present the expression $\frac(\ln(1+x))(x)$ in the following form: $\frac(1)(x)\cdot\ln(1+x)$. Now let's factor $\frac(1)(x)$ into the power of the expression $(1+x)$ and apply the second remarkable limit:
$$ \lim_(x\to\ 0) \frac(\ln(1+x))(x)=\left| \frac(0)(0) \right|= \lim_(x\to\ 0) \left(\frac(1)(x)\cdot\ln(1+x)\right)=\lim_(x\ to\ 0)\ln(1+x)^(\frac(1)(x))=\ln e=1. $$
Once again we have uncertainty of the form $\frac(0)(0)$. We will rely on the formula we have already proven. Since $\log_a t=\frac(\ln t)(\ln a)$, then $\log_a (1+x)=\frac(\ln(1+x))(\ln a)$.
$$ \lim_(x\to\ 0) \frac(\log_a (1+x))(x)=\left| \frac(0)(0) \right|=\lim_(x\to\ 0)\frac(\ln(1+x))( x \ln a)=\frac(1)(\ln a)\ lim_(x\to\ 0)\frac(\ln(1+x))( x)=\frac(1)(\ln a)\cdot 1=\frac(1)(\ln a). $$
Corollary #2
\begin(equation) \lim_(x\to\ 0) \frac(e^x-1)(x)=1\end(equation)Evidence of corollary No. 2: show\hide
Since at $x\to 0$ we have $e^x-1\to 0$, then in the limit under consideration there is an uncertainty of the form $\frac(0)(0)$. To reveal this uncertainty, let us change the variable, denoting $t=e^x-1$. Since $x\to 0$, then $t\to 0$. Next, from the formula $t=e^x-1$ we get: $e^x=1+t$, $x=\ln(1+t)$.
$$ \lim_(x\to\ 0) \frac(e^x-1)(x)=\left| \frac(0)(0) \right|=\left | \begin(aligned) & t=e^x-1;\; t\to 0.\\ & x=\ln(1+t).\end (aligned) \right|= \lim_(t\to 0)\frac(t)(\ln(1+t))= \lim_(t\to 0)\frac(1)(\frac(\ln(1+t))(t))=\frac(1)(1)=1. $$
Once again we have uncertainty of the form $\frac(0)(0)$. We will rely on the formula we have already proven. Since $a^x=e^(x\ln a)$, then:
$$ \lim_(x\to\ 0) \frac(a^(x)-1)(x)=\left| \frac(0)(0) \right|=\lim_(x\to 0)\frac(e^(x\ln a)-1)(x)=\ln a\cdot \lim_(x\to 0 )\frac(e^(x\ln a)-1)(x \ln a)=\ln a \cdot 1=\ln a. $$
Corollary #3
\begin(equation) \lim_(x\to\ 0) \frac((1+x)^\alpha-1)(x)=\alpha \end(equation)Evidence of corollary No. 3: show\hide
Once again we are dealing with uncertainty of the form $\frac(0)(0)$. Since $(1+x)^\alpha=e^(\alpha\ln(1+x))$, we get:
$$ \lim_(x\to\ 0) \frac((1+x)^\alpha-1)(x)= \left| \frac(0)(0) \right|= \lim_(x\to\ 0)\frac(e^(\alpha\ln(1+x))-1)(x)= \lim_(x\to \ 0)\left(\frac(e^(\alpha\ln(1+x))-1)(\alpha\ln(1+x))\cdot \frac(\alpha\ln(1+x) )(x) \right)=\\ =\alpha\lim_(x\to\ 0) \frac(e^(\alpha\ln(1+x))-1)(\alpha\ln(1+x ))\cdot \lim_(x\to\ 0)\frac(\ln(1+x))(x)=\alpha\cdot 1\cdot 1=\alpha. $$
Example No. 1
Calculate the limit $\lim_(x\to\ 0) \frac(e^(9x)-1)(\sin 5x)$.
We have an uncertainty of the form $\frac(0)(0)$. To reveal this uncertainty, we will use the formula. To fit our limit to this formula It should be borne in mind that the expressions in the power of $e$ and in the denominator must coincide. In other words, there is no place for sine in the denominator. The denominator should be $9x$. Additionally, the solution to this example will use the first remarkable limit.
$$ \lim_(x\to\ 0) \frac(e^(9x)-1)(\sin 5x)=\left|\frac(0)(0) \right|=\lim_(x\to\ 0) \left(\frac(e^(9x)-1)(9x)\cdot\frac(9x)(\sin 5x) \right) =\frac(9)(5)\cdot\lim_(x\ to\ 0) \left(\frac(e^(9x)-1)(9x)\cdot\frac(1)(\frac(\sin 5x)(5x)) \right)=\frac(9)( 5)\cdot 1 \cdot 1=\frac(9)(5). $$
Answer: $\lim_(x\to\ 0) \frac(e^(9x)-1)(\sin 5x)=\frac(9)(5)$.
Example No. 2
Calculate the limit $\lim_(x\to\ 0) \frac(\ln\cos x)(x^2)$.
We have an uncertainty of the form $\frac(0)(0)$ (let me remind you that $\ln\cos 0=\ln 1=0$). To reveal this uncertainty, we will use the formula. First, let's take into account that $\cos x=1-2\sin^2 \frac(x)(2)$ (see printout on trigonometric functions). Now $\ln\cos x=\ln\left(1-2\sin^2 \frac(x)(2)\right)$, so in the denominator we should get the expression $-2\sin^2 \frac(x )(2)$ (to fit our example to the formula). In the further solution, the first remarkable limit will be used.
$$ \lim_(x\to\ 0) \frac(\ln\cos x)(x^2)=\left| \frac(0)(0) \right|=\lim_(x\to\ 0) \frac(\ln\left(1-2\sin^2 \frac(x)(2)\right))(x ^2)= \lim_(x\to\ 0) \left(\frac(\ln\left(1-2\sin^2 \frac(x)(2)\right))(-2\sin^2 \frac(x)(2))\cdot\frac(-2\sin^2 \frac(x)(2))(x^2) \right)=\\ =-\frac(1)(2) \lim_(x\to\ 0) \left(\frac(\ln\left(1-2\sin^2 \frac(x)(2)\right))(-2\sin^2 \frac(x )(2))\cdot\left(\frac(\sin\frac(x)(2))(\frac(x)(2))\right)^2 \right)=-\frac(1)( 2)\cdot 1\cdot 1^2=-\frac(1)(2). $$
Answer: $\lim_(x\to\ 0) \frac(\ln\cos x)(x^2)=-\frac(1)(2)$.
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.
Yandex.RTB R-A-339285-1
Let's consider problems in which the ability to calculate the second remarkable limit will be useful.
Example 1
Find the limit lim x → ∞ 1 - 2 x 2 + 1 x 2 + 1 4 .
Solution
Let's substitute the required formula and perform the calculations.
lim x → ∞ 1 - 2 x 2 + 1 x 2 + 1 4 = 1 - 2 ∞ 2 + 1 ∞ 2 + 1 4 = 1 - 0 ∞ = 1 ∞
Our answer turned out to be one to the power of infinity. To determine the solution method, we use the uncertainty table. Let's choose the second remarkable limit and make a change of variables.
t = - x 2 + 1 2 ⇔ x 2 + 1 4 = - t 2
If x → ∞, then t → - ∞.
Let's see what we got after the replacement:
lim x → ∞ 1 - 2 x 2 + 1 x 2 + 1 4 = 1 ∞ = lim x → ∞ 1 + 1 t - 1 2 t = lim t → ∞ 1 + 1 t t - 1 2 = e - 1 2
Answer: lim x → ∞ 1 - 2 x 2 + 1 x 2 + 1 4 = e - 1 2 .
Example 2
Calculate the limit lim x → ∞ x - 1 x + 1 x .
Solution
Let's substitute infinity and get the following.
lim x → ∞ x - 1 x + 1 x = lim x → ∞ 1 - 1 x 1 + 1 x x = 1 - 0 1 + 0 ∞ = 1 ∞
In the answer, we again got the same thing as in the previous problem, therefore, we can again use the second remarkable limit. Next we need to select at the base power function whole part:
x - 1 x + 1 = x + 1 - 2 x + 1 = x + 1 x + 1 - 2 x + 1 = 1 - 2 x + 1
After this, the limit takes on the following form:
lim x → ∞ x - 1 x + 1 x = 1 ∞ = lim x → ∞ 1 - 2 x + 1 x
Replace variables. Let's assume that t = - x + 1 2 ⇒ 2 t = - x - 1 ⇒ x = - 2 t - 1 ; if x → ∞, then t → ∞.
After that, we write down what we got in the original limit:
lim x → ∞ x - 1 x + 1 x = 1 ∞ = lim x → ∞ 1 - 2 x + 1 x = lim x → ∞ 1 + 1 t - 2 t - 1 = = lim x → ∞ 1 + 1 t - 2 t 1 + 1 t - 1 = lim x → ∞ 1 + 1 t - 2 t lim x → ∞ 1 + 1 t - 1 = = lim x → ∞ 1 + 1 t t - 2 1 + 1 ∞ = e - 2 · (1 + 0) - 1 = e - 2
To perform this transformation, we used the basic properties of limits and powers.
Answer: lim x → ∞ x - 1 x + 1 x = e - 2 .
Example 3
Calculate the limit lim x → ∞ x 3 + 1 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 .
Solution
lim x → ∞ x 3 + 1 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 = lim x → ∞ 1 + 1 x 3 1 + 2 x - 1 x 3 3 2 x - 5 x 4 = = 1 + 0 1 + 0 - 0 3 0 - 0 = 1 ∞
After that, we need to transform the function to apply the second great limit. We got the following:
lim x → ∞ x 3 + 1 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 = 1 ∞ = lim x → ∞ x 3 - 2 x 2 - 1 - 2 x 2 + 2 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 = = lim x → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5
lim x → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 = lim x → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 x 3 + 2 x 2 - 1 - 2 x 2 + 2 - 2 x 2 + 2 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 = = lim x → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 x 3 + 2 x 2 - 1 - 2 x 2 + 2 - 2 x 2 + 2 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5
Since we now have the same exponents in the numerator and denominator of the fraction (equal to six), the limit of the fraction at infinity will be equal to the ratio of these coefficients at higher powers.
lim x → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 x 3 + 2 x 2 - 1 - 2 x 2 + 2 - 2 x 2 + 2 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 = = lim x → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 x 3 + 2 x 2 - 1 - 2 x 2 + 2 - 6 2 = lim x → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 x 3 + 2 x 2 - 1 - 2 x 2 + 2 - 3
By substituting t = x 2 + 2 x 2 - 1 - 2 x 2 + 2 we get a second remarkable limit. Means what:
lim x → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 x 3 + 2 x 2 - 1 - 2 x 2 + 2 - 3 = lim x → ∞ 1 + 1 t t - 3 = e - 3
Answer: lim x → ∞ x 3 + 1 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 = e - 3 .
conclusions
Uncertainty 1 ∞, i.e. unity to an infinite power is a power-law uncertainty, therefore, it can be revealed using the rules for finding the limits of exponential power functions.
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A little theory.
Limit of the function at x->x 0
Let the function f(x) be defined on some set X and let the point \(x_0 \in X\) or \(x_0 \notin X\)
Let us take from X a sequence of points different from x 0:
x 1 , x 2 , x 3 , ..., x n , ... (1)
converging to x*. The function values at the points of this sequence also form a numerical sequence
f(x 1), f(x 2), f(x 3), ..., f(x n), ... (2)
and one can raise the question of the existence of its limit.
Definition. The number A is called the limit of the function f(x) at the point x = x 0 (or at x -> x 0), if for any sequence (1) of values of the argument x different from x 0 converging to x 0, the corresponding sequence (2) of values function converges to number A.
$$ \lim_(x\to x_0)( f(x)) = A $$
The function f(x) can have only one limit at the point x 0. This follows from the fact that the sequence
(f(x n)) has only one limit.
There is another definition of the limit of a function.
Definition The number A is called the limit of the function f(x) at the point x = x 0 if for any number \(\varepsilon > 0\) there is a number \(\delta > 0\) such that for all \(x \in X, \; x \neq x_0 \), satisfying the inequality \(|x-x_0| Using logical symbols, this definition can be written as
\((\forall \varepsilon > 0) (\exists \delta > 0) (\forall x \in X, \; x \neq x_0, \; |x-x_0| Note that the inequalities \(x \neq x_0 , \; |x-x_0| The first definition is based on the concept of limit number sequence, which is why it is often called a "sequence language" definition. The second definition is called the definition "in the language \(\varepsilon - \delta \)".
These two definitions of the limit of a function are equivalent and you can use either of them depending on which is more convenient for solving a particular problem.
Note that the definition of the limit of a function “in the language of sequences” is also called the definition of the limit of a function according to Heine, and the definition of the limit of a function “in the language \(\varepsilon - \delta \)” is also called the definition of the limit of a function according to Cauchy.
Limit of the function at x->x 0 - and at x->x 0 +
In what follows, we will use the concepts of one-sided limits of a function, which are defined as follows.
Definition The number A is called the right (left) limit of the function f(x) at the point x 0 if for any sequence (1) converging to x 0, the elements x n of which are greater (less than) x 0, the corresponding sequence (2) converges to A.
Symbolically it is written like this:
$$ \lim_(x \to x_0+) f(x) = A \; \left(\lim_(x \to x_0-) f(x) = A \right) $$
We can give an equivalent definition of one-sided limits of a function “in the language \(\varepsilon - \delta \)”:
Definition a number A is called the right (left) limit of the function f(x) at the point x 0 if for any \(\varepsilon > 0\) there is a \(\delta > 0\) such that for all x satisfying the inequalities \(x_0 Symbolic entries:
