Solution online function limits. Find the limit value of a function or functional sequence at a point, calculate limiting function value at infinity. determine the convergence of a number series and much more can be done thanks to our online service -. We allow you to find function limits online quickly and accurately. You yourself enter the function variable and the limit to which it aspires, our service does all the calculations for you, giving an accurate and simple answer. And for finding the limit online you can enter both numerical series and analytic functions containing constants in a literal expression. In this case, the found function limit will contain these constants as constant arguments in the expression. Our service solves any complex problems of finding limits online, it is enough to specify the function and the point at which it is necessary to calculate function limit. Computing limits online, you can use various methods and rules for solving them, while comparing the result with limit solution online on www.site, which will lead to the successful completion of the task - you will avoid your own mistakes and typos. Or you can completely trust us and use our result in your work, without spending extra effort and time on independent calculations of the function limit. We allow input of limit values such as infinity. You must enter a common term of the numerical sequence and www.site will calculate the value limit online to plus or minus infinity.
One of the basic concepts of mathematical analysis is function limit And sequence limit at a point and at infinity, it is important to be able to correctly solve limits. With our service it will not be difficult. A decision is being made limits online within seconds, the answer is accurate and complete. The study of calculus begins with passage to the limit, limits are used in almost all sections of higher mathematics, so it is useful to have a server at hand for limit solutions online which is the site.
For those who want to learn how to find the limits in this article we will talk about it. We will not delve into the theory, it is usually given in lectures by teachers. So the "boring theory" should be outlined in your notebooks. If this is not the case, then you can read textbooks taken from the library of the educational institution or on other Internet resources.
So, the concept of the limit is quite important in the study of the course of higher mathematics, especially when you come across the integral calculus and understand the relationship between the limit and the integral. In the current material, simple examples will be considered, as well as ways to solve them.
Solution examples
| Example 1 |
| Calculate a) $ \lim_(x \to 0) \frac(1)(x) $; b)$ \lim_(x \to \infty) \frac(1)(x) $ |
| Solution |
|
a) $$ \lim \limits_(x \to 0) \frac(1)(x) = \infty $$ b)$$ \lim_(x \to \infty) \frac(1)(x) = 0 $$ We often get these limits sent to us asking for help to solve. We decided to highlight them as a separate example and explain that these limits simply need to be remembered, as a rule. If you cannot solve your problem, then send it to us. We will provide a detailed solution. You will be able to familiarize yourself with the progress of the calculation and gather information. This will help you get a credit from the teacher in a timely manner! |
| Answer |
| $$ \text(a)) \lim \limits_(x \to \to 0) \frac(1)(x) = \infty \text( b))\lim \limits_(x \to \infty) \frac(1 )(x) = 0 $$ |
What to do with the uncertainty of the form: $ \bigg [\frac(0)(0) \bigg ] $
| Example 3 |
| Solve $ \lim \limits_(x \to -1) \frac(x^2-1)(x+1) $ |
| Solution |
|
As always, we start by substituting the value of $ x $ into the expression under the limit sign. $$ \lim \limits_(x \to -1) \frac(x^2-1)(x+1) = \frac((-1)^2-1)(-1+1)=\frac( 0)(0) $$ What's next? What should be the result? Since this is an uncertainty, this is not yet an answer and we continue the calculation. Since we have a polynomial in the numerators, we decompose it into factors using the familiar formula $$ a^2-b^2=(a-b)(a+b) $$. Remembered? Great! Now go ahead and apply it with the song :) We get that the numerator $ x^2-1=(x-1)(x+1) $ We continue to solve given the above transformation: $$ \lim \limits_(x \to -1)\frac(x^2-1)(x+1) = \lim \limits_(x \to -1)\frac((x-1)(x+ 1))(x+1) = $$ $$ = \lim \limits_(x \to -1)(x-1)=-1-1=-2 $$ |
| Answer |
| $$ \lim \limits_(x \to -1) \frac(x^2-1)(x+1) = -2 $$ |
Let's take the limit in the last two examples to infinity and consider the uncertainty: $ \bigg [\frac(\infty)(\infty) \bigg ] $
| Example 5 |
| Calculate $ \lim \limits_(x \to \infty) \frac(x^2-1)(x+1) $ |
| Solution |
|
$ \lim \limits_(x \to \infty) \frac(x^2-1)(x+1) = \frac(\infty)(\infty) $ What to do? How to be? Do not panic, because the impossible is possible. It is necessary to take out the brackets in both the numerator and the denominator X, and then reduce it. After that, try to calculate the limit. Trying... $$ \lim \limits_(x \to \infty) \frac(x^2-1)(x+1) =\lim \limits_(x \to \infty) \frac(x^2(1-\frac (1)(x^2)))(x(1+\frac(1)(x))) = $$ $$ = \lim \limits_(x \to \infty) \frac(x(1-\frac(1)(x^2)))((1+\frac(1)(x))) = $$ Using the definition from Example 2 and substituting infinity for x, we get: $$ = \frac(\infty(1-\frac(1)(\infty)))((1+\frac(1)(\infty))) = \frac(\infty \cdot 1)(1+ 0) = \frac(\infty)(1) = \infty $$ |
| Answer |
| $$ \lim \limits_(x \to \infty) \frac(x^2-1)(x+1) = \infty $$ |
Algorithm for calculating limits
So, let's briefly summarize the analyzed examples and make an algorithm for solving the limits:
- Substitute point x in the expression following the limit sign. If a certain number is obtained, or infinity, then the limit is completely solved. Otherwise, we have uncertainty: "zero divided by zero" or "infinity divided by infinity" and proceed to the next paragraphs of the instruction.
- To eliminate the uncertainty "zero divide by zero" you need to factorize the numerator and denominator. Reduce similar. Substitute the point x in the expression under the limit sign.
- If the uncertainty is "infinity divided by infinity", then we take out both in the numerator and in the denominator x of the greatest degree. We shorten the x's. We substitute x values from under the limit into the remaining expression.
In this article, you got acquainted with the basics of solving limits, often used in the Calculus course. Of course, these are not all types of problems offered by examiners, but only the simplest limits. We will talk about other types of tasks in future articles, but first you need to learn this lesson in order to move on. We will discuss what to do if there are roots, degrees, we will study infinitesimal equivalent functions, wonderful limits, L'Hopital's rule.
If you can't figure out the limits on your own, don't panic. We are always happy to help!
The first remarkable limit is called the following equality:
\begin(equation)\lim_(\alpha\to(0))\frac(\sin\alpha)(\alpha)=1 \end(equation)
Since for $\alpha\to(0)$ we have $\sin\alpha\to(0)$, we say that the first remarkable limit reveals an indeterminacy of the form $\frac(0)(0)$. Generally speaking, in formula (1), instead of the variable $\alpha$, under the sine sign and in the denominator, any expression can be located, as long as two conditions are met:
- The expressions under the sine sign and in the denominator simultaneously tend to zero, i.e. there is an uncertainty of the form $\frac(0)(0)$.
- 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 #2, #3, #4 and #5 contain solutions with detailed comments. Examples 6-10 contain solutions with little or no comment, as detailed explanations were given in the previous examples. When solving, some trigonometric formulas are used, which can be found.
I note that the presence of trigonometric functions, coupled with the uncertainty of $\frac (0) (0)$, does not mean that the first remarkable limit must be applied. Sometimes simple trigonometric transformations are enough - for example, see.
Example #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 replacement $\alpha=\sin(y)$. Since $\sin(0)=0$, then from the condition $\alpha\to(0)$ we have $y\to(0)$. In addition, there is a neighborhood of zero where $\arcsin\alpha=\arcsin(\sin(y))=y$, 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 $\lim_(\alpha\to(0))\frac(\arcsin\alpha)(\alpha)=1$ is proved.
c) Let's make the replacement $\alpha=\tg(y)$. Since $\tg(0)=0$, the conditions $\alpha\to(0)$ and $y\to(0)$ are equivalent. In addition, there is a neighborhood of zero where $\arctg\alpha=\arctg\tg(y))=y$, therefore, relying 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 $\lim_(\alpha\to(0))\frac(\arctg\alpha)(\alpha)=1$ is proved.
Equalities a), b), c) are often used along with the first remarkable limit.
Example #2
Compute 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 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 can be seen that the expressions under the sine sign and in the denominator are the same (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 #3
Find $\lim_(x\to(0))\frac(\sin(9x))(x)$.
Since $\lim_(x\to(0))\sin(9x)=0$ and $\lim_(x\to(0))x=0$, 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 match. Here it is required 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. Basically, we're missing the $9$ factor in the denominator, which isn't that hard to enter, just multiply the expression in the denominator by $9$. Naturally, to compensate for the multiplication by $9$, you will have to immediately divide by $9$ and divide:
$$ \lim_(x\to(0))\frac(\sin(9x))(x)=\left|\frac(0)(0)\right| =\lim_(x\to(0))\frac(\sin(9x))(9x\cdot\frac(1)(9)) =9\lim_(x\to(0))\frac(\sin (9x))(9x) $$
Now the expressions in the denominator and under the sine sign are the same. Both conditions for the limit $\lim_(x\to(0))\frac(\sin(9x))(9x)$ are satisfied. Hence $\lim_(x\to(0))\frac(\sin(9x))(9x)=1$. 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 #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 an indeterminacy of the form $\frac(0)(0)$. However, the form of the first remarkable limit is broken. A numerator containing $\sin(5x)$ requires $5x$ in the denominator. 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 $\frac(5)(8)$ out of 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 #5
Find $\lim_(x\to(0))\frac(\cos(5x)-\cos^3(5x))(x^2)$.
Since $\lim_(x\to(0))(\cos(5x)-\cos^3(5x))=1-1=0$ (recall that $\cos(0)=1$) and $\lim_(x\to(0))x^2=0$, then we are dealing with an indeterminacy of the form $\frac(0)(0)$. However, in order to apply the first wonderful limit, you should get rid of the cosine in the numerator by going to sines (in order to then apply the formula) or tangents (in order to then apply the formula). You can do this 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 wonderful 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 considered limit:
$$ \lim_(x\to(0))\left(\cos(5x)\cdot\frac(\sin^2(5x))(x^2)\right) =\lim_(x\to(0 ))\left(25\cos(5x)\cdot\left(\frac(\sin(5x))(5x)\right)^2\right)=\\ =25\cdot\lim_(x\to( 0))\cos(5x)\cdot\lim_(x\to(0))\left(\frac(\sin(5x))(5x)\right)^2 =25\cdot(1)\cdot( 1^2) =25. $$
Answer: $\lim_(x\to(0))\frac(\cos(5x)-\cos^3(5x))(x^2)=25$.
Example #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 the uncertainty of $\frac(0)(0)$. Let's open 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 in the given limit to sines, 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 #7
Calculate limit $\lim_(x\to(0))\frac(\cos(\alpha(x))-\cos(\beta(x)))(x^2)$ given $\alpha\neq\ beta$.
Detailed explanations were given earlier, but here we simply note that again there is an indeterminacy of $\frac(0)(0)$. 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 the above 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 #8
Find the limit $\lim_(x\to(0))\frac(\tg(x)-\sin(x))(x^3)$.
Since $\lim_(x\to(0))(\tg(x)-\sin(x))=0$ (recall that $\sin(0)=\tg(0)=0$) and $\lim_(x\to(0))x^3=0$, then here we are dealing with an indeterminacy of the form $\frac(0)(0)$. Let's break it down like this:
$$ \lim_(x\to(0))\frac(\tg(x)-\sin(x))(x^3)=\left|\frac(0)(0)\right| =\lim_(x\to(0))\frac(\frac(\sin(x))(\cos(x))-\sin(x))(x^3) =\lim_(x\to( 0))\frac(\sin(x)\cdot\left(\frac(1)(\cos(x))-1\right))(x^3) =\lim_(x\to(0)) \frac(\sin(x)\cdot\left(1-\cos(x)\right))(x^3\cdot\cos(x))=\\ =\lim_(x\to(0)) \frac(\sin(x)\cdot(2)\sin^2\frac(x)(2))(x^3\cdot\cos(x)) =\frac(1)(2)\cdot\ lim_(x\to(0))\left(\frac(\sin(x))(x)\cdot\left(\frac(\sin\frac(x)(2))(\frac(x)( 2))\right)^2\cdot\frac(1)(\cos(x))\right) =\frac(1)(2)\cdot(1)\cdot(1^2)\cdot(1 ) =\frac(1)(2). $$
Answer: $\lim_(x\to(0))\frac(\tg(x)-\sin(x))(x^3)=\frac(1)(2)$.
Example #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 an indeterminacy of the form $\frac(0)(0)$. Before proceeding to its expansion, it is convenient to change the variable in such a way that the new variable tends to zero (note that the variable $\alpha \to 0$ in the formulas). The easiest way is to introduce the variable $t=x-3$. However, for the 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 substitutions are applicable in this case, just the second substitution 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 #10
Find the limit $\lim_(x\to\frac(\pi)(2))\frac(1-\sin(x))(\left(\frac(\pi)(2)-x\right)^2 )$.
Again we are dealing with the uncertainty of $\frac(0)(0)$. Before proceeding to its expansion, it is convenient to make a variable change 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 #11
Find 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 do not have to use the first wonderful limit. Please note: in both the first and second limits, there are only trigonometric functions and numbers. Often, in examples of this kind, it is possible to simplify the expression located under the limit sign. In this case, after the mentioned simplification and reduction of some factors, the uncertainty disappears. I gave this example with only one purpose: to show that the presence of trigonometric functions under the limit sign does not necessarily mean the application of the first remarkable limit.
Since $\lim_(x\to\frac(\pi)(2))(1-\sin(x))=0$ (recall that $\sin\frac(\pi)(2)=1$ ) and $\lim_(x\to\frac(\pi)(2))\cos^2x=0$ (recall that $\cos\frac(\pi)(2)=0$), then we have dealing with uncertainty of the form $\frac(0)(0)$. However, this does not mean at all that we need to use the first remarkable limit. To reveal the uncertainty, it suffices 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 of 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 expressions in the numerator and denominator. The purpose of our actions: write the sum in the numerator and denominator as a product. By the way, it is often convenient to change a variable within a similar form so 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 the variable, although it is easy to implement the replacement of the variable $t=x-\frac(2\pi)(3)$ if desired.
$$ \lim_(x\to\frac(2\pi)(3))\frac(\tg(x)+\sqrt(3))(2\cos(x)+1) =\lim_(x\ to\frac(2\pi)(3))\frac(\tg(x)+\sqrt(3))(2\cdot\left(\cos(x)+\frac(1)(2)\right )) =\lim_(x\to\frac(2\pi)(3))\frac(\tg(x)-\tg\frac(2\pi)(3))(2\cdot\left(\ cos(x)-\cos\frac(2\pi)(3)\right))=\\ =\lim_(x\to\frac(2\pi)(3))\frac(\frac(\sin \left(x-\frac(2\pi)(3)\right))(\cos(x)\cos\frac(2\pi)(3)))(-4\sin\frac(x+\frac (2\pi)(3))(2)\sin\frac(x-\frac(2\pi)(3))(2)) =\lim_(x\to\frac(2\pi)(3 ))\frac(\sin\left(x-\frac(2\pi)(3)\right))(-4\sin\frac(x+\frac(2\pi)(3))(2)\ sin\frac(x-\frac(2\pi)(3))(2)\cos(x)\cos\frac(2\pi)(3))=\\ =\lim_(x\to\frac (2\pi)(3))\frac(2\sin\frac(x-\frac(2\pi)(3))(2)\cos\frac(x-\frac(2\pi)(3 ))(2))(-4\sin\frac(x+\frac(2\pi)(3))(2)\sin\frac(x-\frac(2\pi)(3))(2) \cos(x)\cos\frac(2\pi)(3)) =\lim_(x\to\frac(2\pi)(3))\frac(\cos\frac(x-\frac(2 \pi)(3))(2))(-2\sin\frac(x+\frac(2\pi)(3))(2)\cos(x)\cos\frac(2\pi)(3 ))=\\ =\frac(1)(-2\cdot\frac(\sqrt(3))(2)\cdot\left(-\frac(1)(2)\right)\cdot\left( -\frac(1)(2)\right)) =-\frac(4)(\sqrt(3)). $$
As you can see, we didn't have to apply the first wonderful limit. Of course, this can be done if desired (see note below), but it is not necessary.
What would be 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: $\lim_(x\to\frac(\pi)(2))\frac(1-\sin(x))(\cos^2x)=\frac(1)(2)$, $\lim_( x\to\frac(2\pi)(3))\frac(\tg(x)+\sqrt(3))(2\cos(x)+1)=-\frac(4)(\sqrt( 3))$.
Limits give all students of mathematics a lot of trouble. To solve the limit, sometimes you have to use a lot of tricks and choose from a variety of solutions exactly the one that is suitable for a particular example.
In this article, we will not help you understand the limits of your abilities or comprehend the limits of control, but we will try to answer the question: how to understand the limits in higher mathematics? Understanding comes with experience, so at the same time we will give some detailed examples of solving limits with explanations.
The concept of a limit in mathematics
The first question is: what is the limit and the limit of what? We can talk about the limits of numerical sequences and functions. We are interested in the concept of the limit of a function, since it is with them that students most often encounter. But first, the most general definition of a limit:
Let's say there is some variable. If this value in the process of change indefinitely approaches a certain number a , That a is the limit of this value.
For a function defined in some interval f(x)=y the limit is the number A , to which the function tends when X tending to a certain point A . Dot A belongs to the interval on which the function is defined.
It sounds cumbersome, but it is written very simply:
Lim- from English limit- limit.
There is also a geometric explanation for the definition of the limit, but here we will not go into theory, since we are more interested in the practical than the theoretical side of the issue. When we say that X tends to some value, this means that the variable does not take on the value of a number, but approaches it infinitely close.
Let's take a concrete example. The challenge is to find the limit.
To solve this example, we substitute the value x=3 into a function. We get:
By the way, if you are interested, read a separate article on this topic.
In the examples X can tend to any value. It can be any number or infinity. Here is an example when X tends to infinity:
It is intuitively clear that the larger the number in the denominator, the smaller the value will be taken by the function. So, with unlimited growth X meaning 1/x will decrease and approach zero.
As you can see, in order to solve the limit, you just need to substitute the value to strive for into the function X . However, this is the simplest case. Often finding the limit is not so obvious. Within the limits there are uncertainties of type 0/0 or infinity/infinity . What to do in such cases? Use tricks!
Uncertainties within
Uncertainty of the form infinity/infinity
Let there be a limit:
If we try to substitute infinity into the function, we will get infinity both in the numerator and in the denominator. In general, it is worth saying that there is a certain element of art in resolving such uncertainties: one must notice how a function can be transformed in such a way that the uncertainty is gone. In our case, we divide the numerator and denominator by X in senior degree. What will happen?
From the example already considered above, we know that terms containing x in the denominator will tend to zero. Then the solution to the limit is:
To uncover type ambiguities infinity/infinity divide the numerator and denominator by X to the highest degree.
By the way! For our readers there is now a 10% discount on
Another type of uncertainty: 0/0
As always, substitution into the value function x=-1 gives 0 in the numerator and denominator. Look a little more closely and you will notice that we have a quadratic equation in the numerator. Let's find the roots and write:
Let's reduce and get:
So, if you encounter type ambiguity 0/0 - factorize the numerator and denominator.
To make it easier for you to solve examples, here is a table with the limits of some functions:
L'Hopital's rule within
Another powerful way to eliminate both types of uncertainties. What is the essence of the method?
If there is uncertainty in the limit, we take the derivative of the numerator and denominator until the uncertainty disappears.
Visually, L'Hopital's rule looks like this:
Important point : the limit, in which the derivatives of the numerator and denominator are instead of the numerator and denominator, must exist.
And now a real example:
There is a typical uncertainty 0/0 . Take the derivatives of the numerator and denominator:
Voila, the uncertainty is eliminated quickly and elegantly.
We hope that you will be able to put this information to good use in practice and find the answer to the question "how to solve limits in higher mathematics". If you need to calculate the limit of a sequence or the limit of a function at a point, but there is no time for this work from the word “absolutely”, contact professional student service for a fast and detailed solution.