An online limit calculator on the site for the full consolidation of the material covered by students and schoolchildren and training their practical skills. How to use the limit calculator online on our resource? This is done even very easily, you just need to enter the original function in the existing field, select the required limit value for the variable from the selector and click on the "Solution" button. If at some point you need to calculate the limit value, then you need to enter the value of this very point - either numeric or symbolic. The online limit calculator will help you find the limit value at a given point, the limit in the function definition interval, and this value, where the value of the function under study rushes when its argument tends to a given point, is the solution to the limit. 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Here the limits of functions, as well as the solution of these limits, are studied only at points that are limiting for the domain of definition of functions, knowing that in each neighborhood of such a point there are points from the domain of definition of this function. This allows us to talk about the tendency of a variable function to a given point. If there is a limit at some point of the function domain and the online limit calculator gives a detailed limit solution of the function at this point, then the function is continuous at this point. Let our online limit calculator with a solution give some positive result, and we will check it on other sites. This can prove the quality of our resource, and, as many already know, it is at its best and deserves the highest praise. Along with this, there is the possibility of online calculator limits with a detailed solution to study and independently, but under the close supervision of a professional teacher. Often this action will lead to the expected results. All students just dream that the online limit calculator with the solution would describe in detail their difficult task, given by the teacher at the beginning of the semester. But it's not so simple. You must first study the theory, and then use the free calculator. Like the online limits, the calculator will give you the details of the entries you need, and you will be satisfied with the result. But the limit point of the domain of definition may not belong to this very domain of definition, and this is proved by a detailed calculation by the online limit calculator. Example: we can consider the limit of a function at the ends of an open segment on which our function is defined. In this case, the boundaries of the segment themselves are not included in the domain of definition. In this sense, the system of neighborhoods of this point is a special case of such a base of subsets. The online limit calculator with a detailed solution is produced in real time and formulas are applied for it in a given explicit analytical form. The limit of a function using the online limit calculator with a detailed solution is a generalization of the concept of the limit of a sequence: initially, the limit of a function at a point was understood to be the limit of a sequence of elements of the range of a function composed of images of points of a sequence of elements of the domain of a function converging to a given point (the limit at which is considered) ; if such a limit exists, then the function is said to converge to the specified value; if such a limit does not exist, then the function is said to diverge. Generally speaking, the theory of passage to the limit is the basic concept of all mathematical analysis. Everything is based precisely on limit transitions, that is, a detailed solution of the limits is the basis of the science of mathematical analysis, and the online limit calculator lays the foundation for student learning. The online limit calculator with a detailed solution on the site is a unique service for getting an accurate and instant answer in real time. Not infrequently, or rather very often, students immediately have difficulties in solving the limits during the initial study of mathematical analysis. We guarantee that solving the limit calculator online on our service is a guarantee of accuracy and getting a high-quality answer. You will receive an answer to a detailed solution of the limit with a calculator in a matter of seconds, you can even say instantly. If you specify incorrect data, that is, characters that are not allowed by the system, it's okay, the service will automatically inform you about an error. 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As you know, to find the sum of a numerical sequence, you only need to correctly express the partial sum of this sequence, and then everything is simple using our free site service, since calculating the limit using our online limit calculator from a partial sum will be the final sum of the numeric sequence. A detailed solution with a limit calculator online using the site service provides students with a way to see the progress of solving problems, which makes understanding the theory of limits easy and accessible to almost everyone. Stay focused and don't let wrong actions get you into trouble with bad grades. Like any detailed solution with an online service limit calculator, the problem will be presented in a convenient and understandable form, with a detailed solution, in compliance with all the rules and regulations for obtaining a solution. At the same time, you can save time and money, since we ask absolutely nothing for it . 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It must contain the variable "x", otherwise the whole function will be treated by the system as a constant. Next, check the limit value if you specified a given point or a symbolic value. It should also contain only Latin letters - this is important! Then you can try again to find a detailed solution of the limits online on our excellent service, and use the result. As soon as they say that the limits of the online solution in detail are very difficult - do not believe, and most importantly, do not panic, everything is permitted within the framework of the training course. We recommend that you, without panic, devote just a few minutes to our service and check the given exercise. If, nevertheless, the limits of the online solution cannot be solved in detail, then you made a typo, because otherwise the site solves almost any problem without much difficulty. But there is no need to think that you can get the desired result immediately without labor and effort. 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In reality, it will be that the online limit calculator with the solution in detail will begin to proportionally represent all the steps of a step-by-step calculation faster.
Concepts of limits of sequences and functions. When it is required to find the limit of a sequence, it is written as follows: lim xn=a. In such a sequence of sequences, xn tends to a, and n tends to infinity. A sequence is usually represented as a series, for example:
x1, x2, x3...,xm,...,xn... .
Sequences are divided into ascending and descending. For example:
xn=n^2 - increasing sequence
yn=1/n - sequence
So, for example, the limit of the sequence xn=1/n^ :
lim1/n^2=0
x→∞
This limit is zero because n→∞ and the sequence 1/n^2 tends to zero.
Usually the variable x tends to a finite limit a, moreover, x is constantly approaching a, and the value of a is constant. This is written as follows: limx = a, while n can also tend to both zero and infinity. There are infinite functions, for them the limit tends to infinity. In other cases, when, for example, the function of slowing down the train, it is possible for a limit tending to zero.
Limits have a number of properties. As a rule, any function has only one limit. This is the main property of the limit. Others are listed below:
* The sum limit is equal to the sum of the limits:
lim(x+y)=limx+limy
* The limit of the product is equal to the product of the limits:
lim(xy)=limx*limy
* The limit of the quotient is equal to the quotient of the limits:
lim(x/y)=lim x/lim y
* The constant factor is taken out of the limit sign:
lim(Cx)=C lim x
Given a function 1 /x where x →∞, its limit is zero. If x→0, then the limit of such a function is equal to ∞.
For trigonometric functions there are of these rules. Since the sin x function always tends to one as it approaches zero, the identity holds for it:
lim sin x/x=1
In a number of functions, when calculating the limits of which uncertainty arises - a situation in which the limit cannot be calculated. The only way out of this situation is L'Hopital. There are two types of uncertainties:
* uncertainty of the form 0/0
* uncertainty of the form ∞/∞
For example, given a limit of the following form: lim f(x)/l(x), moreover, f(x0)=l(x0)=0. In this case, there is an uncertainty of the form 0/0. To solve such a problem, both functions are differentiated, after which the limit of the result is found. For uncertainties of the form 0/0, the limit is:
lim f(x)/l(x)=lim f"(x)/l"(x) (for x→0)
The same rule is true for uncertainties of the type ∞/∞. But in this case, the following equality is true: f(x)=l(x)=∞
With the help of L'Hopital's rule, one can find the values of any limits in which uncertainties appear. Mandatory condition for
volume - the absence of errors in finding derivatives. So, for example, the derivative of the function (x^2)" is equal to 2x. From this we can conclude that:
f"(x)=nx^(n-1)
Function limit- number a will be the limit of some variable value if, in the process of its change, this variable approaches indefinitely a.
Or in other words, the number A is the limit of the function y=f(x) at the point x0, if for any sequence of points from the domain of definition of the function , not equal to x0, and which converges to the point x 0 (lim x n = x0), the sequence of corresponding values of the function converges to the number A.
Graph of a function whose limit with an argument that tends to infinity is L:
Meaning BUT is limit (limit value) of the function f(x) at the point x0 if for any sequence of points 
, which converges to x0, but which does not contain x0 as one of its elements (i.e. in the punctured neighborhood x0), the sequence of function values 
converges to A.
The limit of a function according to Cauchy.
Meaning A will be function limit f(x) at the point x0 if for any forward taken non-negative number ε a non-negative corresponding number will be found δ = δ(ε) such that for each argument x, satisfying the condition 0 < | x - x0 | < δ , the inequality | f(x) A |< ε .
It will be very simple if you understand the essence of the limit and the basic rules for finding it. That the limit of the function f(x) at x aspiring to a equals A, is written like this:
Moreover, the value to which the variable tends x, can be not only a number, but also infinity (∞), sometimes +∞ or -∞, or there may be no limit at all.
To understand how find the limits of a function, it is best to see examples of solutions.
We need to find the limits of the function f(x) = 1/x at:
x→ 2, x→ 0, x→ ∞.
Let's find the solution of the first limit. To do this, you can simply substitute x the number to which it aspires, i.e. 2, we get:
Find the second limit of the function. Here, substitute in pure form 0 instead of x it is impossible, because cannot be divided by 0. But we can take values close to zero, for example, 0.01; 0.001; 0.0001; 0.00001 and so on, with the value of the function f(x) will increase: 100; 1000; 10000; 100000 and so on. Thus, it can be understood that when x→ 0 the value of the function that is under the limit sign will increase indefinitely, i.e. strive for infinity. Which means:
Regarding the third limit. The same situation as in the previous case, it is impossible to substitute ∞ in its purest form. We need to consider the case of unlimited increase x. We alternately substitute 1000; 10000; 100000 and so on, we have that the value of the function f(x) = 1/x will decrease: 0.001; 0.0001; 0.00001; and so on, tending to zero. That's why:
It is necessary to calculate the limit of the function 
Starting to solve the second example, we see the uncertainty. From here we find the highest degree of the numerator and denominator - this is x 3, we take it out of brackets in the numerator and denominator and then reduce it by it:
Answer 
The first step in finding this limit, substitute the value 1 instead of x, resulting in the uncertainty . To solve it, we decompose the numerator into factors , we will do this by finding the roots of the quadratic equation x 2 + 2x - 3:
D \u003d 2 2 - 4 * 1 * (-3) \u003d 4 +12 \u003d 16→ √ D=√16 = 4
x 1,2 = (-2± 4) / 2→ x 1 \u003d -3;x2= 1.
So the numerator would be:
Answer 
This is the definition of its specific value or a specific area where the function falls, which is limited by the limit.
To decide the limits, follow the rules:
Having understood the essence and main limit decision rules, you will get a basic understanding of how to solve them.
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 , then 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, and there is no time for this work from the word “absolutely”, contact a professional student service for a quick and detailed solution.
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 $\lim_(\alpha\to(0))\frac(\sin\alpha)(\alpha)=1$ , then:
$$ \lim_(\alpha\to(0))\frac(\sin(\alpha))(\alpha\cos(\alpha)) =\frac(\displaystyle\lim_(\alpha\to(0)) \frac(\sin(\alpha))(\alpha))(\displaystyle\lim_(\alpha\to(0))\cos(\alpha)) =\frac(1)(1) =1. $$
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. performed. 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., performed. 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: $\lim_(x\to\frac(\pi)(2))\frac(1-\sin(x))(\cos^2x)=\frac(1)(2)$, $\lim_( x\to\frac(2\pi)(3))\frac(\tg(x)+\sqrt(3))(2\cos(x)+1)=-\frac(4)(\sqrt( 3))$.
