Limits give all mathematics students a lot of trouble. To solve a limit, sometimes you have to use a lot of tricks and choose from a variety of solution methods exactly the one that is suitable for a particular example.
In this article we will not help you understand the limits of your capabilities or comprehend the limits of control, but we will try to answer the question: how to understand limits in higher mathematics? Understanding comes with experience, so at the same time we will give a few detailed examples solutions of limits with explanations.
The concept of limit in mathematics
The first question is: what is this 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 this is what students most often encounter. But first - the most general definition limit:
Let's say there is some variable value. If this value in the process of change unlimitedly approaches a certain number a , That a – the limit of this value.
For a function defined in a certain interval f(x)=y such a number is called a limit A , which the function tends to 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 determining the limit, but here we will not delve into the theory, since we are more interested in the practical rather 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 give specific example. The task 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's an example when X tends to infinity:
It’s intuitively clear what’s what larger number in the denominator, the smaller the value the function will take. So, with unlimited growth X meaning 1/x will decrease and approach zero.
As you can see, 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 the type 0/0 or infinity/infinity . What to do in such cases? Resort to 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 in both the numerator and the denominator. In general, it is worth saying that there is a certain element of art in resolving such uncertainties: you need to notice how you can transform the function in such a way that the uncertainty goes away. In our case, we divide the numerator and denominator by X in the senior degree. What will happen?
From the example already discussed above, we know that terms containing x in the denominator will tend to zero. Then the solution to the limit is:
To resolve type uncertainties 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, substituting values into the 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 are faced with type uncertainty 0/0 – factor the numerator and denominator.
To make it easier for you to solve examples, we present a table with the limits of some functions:
L'Hopital's rule within
Another powerful way, allowing to eliminate uncertainties of both types. What is the essence of the method?
If there is uncertainty in the limit, take the derivative of the numerator and denominator until the uncertainty disappears.
L'Hopital's rule looks like this:
Important point : the limit in which the derivatives of the numerator and denominator stand instead of the numerator and denominator must exist.
And now - a real example:
There is typical uncertainty 0/0 . Let's take the derivatives of the numerator and denominator:
Voila, uncertainty is resolved quickly and elegantly.
We hope that you will be able to usefully apply this information 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 absolutely no time for this work, contact a professional student service for a quick and detailed solution.
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:
- The expressions under the sine sign and in the denominator simultaneously tend to zero, i.e. there is uncertainty of the form $\frac(0)(0)$.
- The expressions under the sine sign and in the denominator are the same.
Corollaries from the first one are also often used. wonderful limit:
\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 required 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: $\lim_(x\to\frac(\pi)(2))\frac(1-\sin(x))(\cos^2x)=\frac(1)(2)$, $\lim_( x\to\frac(2\pi)(3))\frac(\tg(x)+\sqrt(3))(2\cos(x)+1)=-\frac(4)(\sqrt( 3))$.
Function limit- number a will be the limit of some variable quantity if, in the process of its change, this variable quantity indefinitely approaches a.
Or in other words, the number A is the limit of the function y = f(x) at the point x 0, if for any sequence of points from the domain of definition of the function , not equal x 0, and which converges to the point x 0 (lim x n = x0), the sequence of corresponding function values converges to the number A.
The graph of a function whose limit, given an argument that tends to infinity, is equal to L:
Meaning A is limit (limit value) of the function f(x) at the point x 0 in case for any sequence of points 
, which converges to x 0, but which does not contain x 0 as one of its elements (i.e. in the punctured vicinity x 0), sequence of function values 
converges to A.
Limit of a Cauchy function.
Meaning A will be limit of the function f(x) at the point x 0 if for any non-negative number taken in advance ε the corresponding non-negative number will be found δ = δ(ε) such that for each argument x, satisfying the condition 0 < | x - x0 | < δ , the inequality will be satisfied | f(x)A |< ε .
It will be very simple if you understand the essence of the limit and the basic rules for finding it. What is the limit of the function f (x) at x striving for 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 look at examples of solutions.
It is necessary to find the limits of the function f (x) = 1/x at:
x→ 2, x→ 0, x→ ∞.
Let's find a solution to the first limit. To do this, you can simply substitute x the number it tends to, i.e. 2, we get:
Let's find the second limit of the function. Substitute here in pure form 0 instead x it is impossible, because You cannot divide by 0. But we can take values close to zero, for example, 0.01; 0.001; 0.0001; 0.00001 and so on, and the value of the function f (x) will increase: 100; 1000; 10000; 100,000 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 without limit, i.e. strive towards 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 substitute 1000 one by one; 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 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:
Answer 
The first step in finding this limit, substitute the value 1 instead x, resulting in uncertainty. To solve it, let’s factorize the numerator and do this using the method of finding roots quadratic equation x 2 + 2x - 3:
D = 2 2 - 4*1*(-3) = 4 +12 = 16→ √ D=√16 = 4
x 1.2 = (-2±4)/2→ x 1 = -3;x 2= 1.
So the numerator will be:
Answer 
This is the definition of its specific value or a certain area where the function falls, which is limited by the limit.
To solve limits, follow the rules:
Having understood the essence and main rules for solving the limit, You'll get basic concept about how to solve them.
Function y = f (x) is a law (rule) according to which each element x of the set X is associated with one and only one element y of the set Y.
Element x ∈ X called function argument or independent variable.
Element y ∈ Y called function value or dependent variable.
The set X is called domain of the function.
Set of elements y ∈ Y, which have preimages in the set X, is called area or set of function values.
The actual function is called limited from above (from below), if there is a number M such that the inequality holds for all:
.
The number function is called limited, if there is a number M such that for all:
.
Top edge or exact upper bound A real function is called the smallest number that limits its range of values from above. That is, this is a number s for which, for everyone and for any, there is an argument whose function value exceeds s′: .
The upper bound of a function can be denoted as follows:
.
Respectively bottom edge or exact lower limit A real function is called the largest number that limits its range of values from below. That is, this is a number i for which, for everyone and for any, there is an argument whose function value is less than i′: .
The infimum of a function can be denoted as follows:
.
Determining the limit of a function
Determination of the limit of a function according to Cauchy
Finite limits of function at end points
Let the function be defined in some neighborhood of the end point, with the possible exception of the point itself. at a point if for any there is such a thing, depending on , that for all x for which , the inequality holds
.
The limit of a function is denoted as follows:
.
Or at .
Using the logical symbols of existence and universality, the definition of the limit of a function can be written as follows:
.
One-sided limits.
Left limit at a point (left-sided limit):
.
Right limit at a point (right-hand limit):
.
The left and right limits are often denoted as follows:
;
.
Finite limits of a function at points at infinity
Limits at points at infinity are determined in a similar way.
.
.
.
They are often referred to as:
;
;
.
Using the concept of neighborhood of a point
If we introduce the concept of a punctured neighborhood of a point, then we can give a unified definition of the finite limit of a function at finite and infinitely distant points:
.
Here for endpoints
;
;
.
Any neighborhood of points at infinity is punctured:
;
;
.
Infinite Function Limits
Definition
Let the function be defined in some punctured neighborhood of a point (finite or at infinity). Limit of function f (x) as x → x 0
equals infinity, if for anyone, arbitrarily large number M > 0
, there is a number δ M > 0
, depending on M, that for all x belonging to the punctured δ M - neighborhood of the point: , the following inequality holds:
.
The infinite limit is denoted as follows:
.
Or at .
Using the logical symbols of existence and universality, the definition of the infinite limit of a function can be written as follows:
.
You can also introduce definitions of infinite limits of certain signs equal to and :
.
.
Universal definition of the limit of a function
Using the concept of a neighborhood of a point, we can give a universal definition of the finite and infinite limit of a function, applicable both for finite (two-sided and one-sided) and infinitely distant points:
.
Determination of the limit of a function according to Heine
Let the function be defined on some set X:.
The number a is called the limit of the function at point:
,
if for any sequence converging to x 0
:
,
whose elements belong to the set X: ,
.
Let us write this definition using the logical symbols of existence and universality:
.
If we take the left-sided neighborhood of the point x as a set X 0 , then we obtain the definition of the left limit. If it is right-handed, then we get the definition of the right limit. If we take the neighborhood of a point at infinity as a set X, we obtain the definition of the limit of a function at infinity.
Theorem
The Cauchy and Heine definitions of the limit of a function are equivalent.
Proof
Properties and theorems of the limit of a function
Further, we assume that the functions under consideration are defined in the corresponding neighborhood of the point, which is a finite number or one of the symbols: . It can also be a one-sided limit point, that is, have the form or . The neighborhood is two-sided for a two-sided limit and one-sided for a one-sided limit.
Basic properties
If the values of the function f (x) change (or make undefined) a finite number of points x 1, x 2, x 3, ... x n, then this change will not affect the existence and value of the limit of the function at an arbitrary point x 0 .
If there is a finite limit, then there is a punctured neighborhood of the point x 0
, on which the function f (x) limited:
.
Let the function have at point x 0
finite non-zero limit:
.
Then, for any number c from the interval , there is such a punctured neighborhood of the point x 0
, what for ,
, If ;
, If .
If, on some punctured neighborhood of the point, , is a constant, then .
If there are finite limits and and on some punctured neighborhood of the point x 0
,
That .
If , and on some neighborhood of the point
,
That .
In particular, if in some neighborhood of a point
,
then if , then and ;
if , then and .
If on some punctured neighborhood of a point x 0
:
,
and there are finite (or infinite of a certain sign) equal limits:
, That
.
Proofs of the main properties are given on the page
"Basic properties of the limits of a function."
Arithmetic properties of the limit of a function
Let the functions and be defined in some punctured neighborhood of the point . And let there be finite limits:
And .
And let C be a constant, that is given number. Then
;
;
;
, If .
If, then.
Proofs of arithmetic properties are given on the page
"Arithmetic properties of the limits of a function".
Cauchy criterion for the existence of a limit of a function
Theorem
In order for a function defined on some punctured neighborhood of a finite or at infinity point x 0
, had a finite limit at this point, it is necessary and sufficient that for any ε > 0
there was such a punctured neighborhood of the point x 0
, that for any points and from this neighborhood, the following inequality holds:
.
Limit of a complex function
Theorem on the limit of a complex function
Let the function have a limit and map a punctured neighborhood of a point onto a punctured neighborhood of a point. Let the function be defined on this neighborhood and have a limit on it.
Here are the final or infinitely distant points: . Neighborhoods and their corresponding limits can be either two-sided or one-sided.
Then there is a limit of a complex function and it is equal to:
.
The limit theorem of a complex function is applied when the function is not defined at a point or has a value different from the limit. To apply this theorem, there must be a punctured neighborhood of the point where the set of values of the function does not contain the point:
.
If the function is continuous at point , then the limit sign can be applied to the argument of the continuous function:
.
The following is a theorem corresponding to this case.
Theorem on the limit of a continuous function of a function
Let there be a limit of the function g (t) as t → t 0
, and it is equal to x 0
:
.
Here is point t 0
can be finite or infinitely distant: .
And let the function f (x) is continuous at point x 0
.
Then there is a limit of the complex function f (g(t)), and it is equal to f (x0):
.
Proofs of the theorems are given on the page
"Limit and continuity of a complex function".
Infinitesimal and infinitely large functions
Infinitesimal functions
Definition
A function is said to be infinitesimal if
.
Sum, difference and product of a finite number of infinitesimal functions at is an infinitesimal function at .
Product of a function bounded on some punctured neighborhood of the point , to an infinitesimal at is an infinitesimal function at .
In order for a function to have a finite limit, it is necessary and sufficient that
,
where is an infinitesimal function at .
"Properties of infinitesimal functions".
Infinitely large functions
Definition
A function is said to be infinitely large if
.
The sum or difference of a bounded function, on some punctured neighborhood of the point , and an infinitely large function at is an infinitely large function at .
If the function is infinitely large for , and the function is bounded on some punctured neighborhood of the point , then
.
If the function , on some punctured neighborhood of the point , satisfies the inequality:
,
and the function is infinitesimal at:
, and (on some punctured neighborhood of the point), then
.
Proofs of the properties are presented in section
"Properties of infinitely large functions".
Relationship between infinitely large and infinitesimal functions
From the two previous properties follows the connection between infinitely large and infinitesimal functions.
If a function is infinitely large at , then the function is infinitesimal at .
If a function is infinitesimal for , and , then the function is infinitely large for .
The relationship between an infinitesimal and an infinitely large function can be expressed symbolically:
,
.
If an infinitesimal function has a certain sign at , that is, it is positive (or negative) on some punctured neighborhood of the point , then this fact can be expressed as follows:
.
In the same way, if an infinitely large function has a certain sign at , then they write:
.
Then the symbolic connection between infinitely small and infinitely large functions can be supplemented with the following relations:
,
,
,
.
Additional formulas relating infinity symbols can be found on the page
"Points at infinity and their properties."
Limits of monotonic functions
Definition
Function defined on some set real numbers X is called strictly increasing, if for all such that the following inequality holds:
.
Accordingly, for strictly decreasing function the following inequality holds:
.
For non-decreasing:
.
For non-increasing:
.
It follows that a strictly increasing function is also non-decreasing. A strictly decreasing function is also non-increasing.
The function is called monotonous, if it is non-decreasing or non-increasing.
Theorem
Let the function not decrease on the interval where .
If it is bounded above by the number M: then there is a finite limit. If not limited from above, then .
If it is limited from below by the number m: then there is a finite limit. If not limited from below, then .
If points a and b are at infinity, then in the expressions the limit signs mean that .
This theorem can be formulated more compactly.
Let the function not decrease on the interval where . Then there are one-sided limits at points a and b:
;
.
A similar theorem for a non-increasing function.
Let the function not increase on the interval where . Then there are one-sided limits:
;
.
The proof of the theorem is presented on the page
"Limits of monotonic functions".
References:
L.D. Kudryavtsev. Course of mathematical analysis. Volume 1. Moscow, 2003.
CM. Nikolsky. Course of mathematical analysis. Volume 1. Moscow, 1983.
