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 $\sin\alpha\to(0)$, 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 remarkable limit are also often used:
\begin(equation) \lim_(\alpha\to(0))\frac(\tg\alpha)(\alpha)=1 \end(equation) \begin(equation) \lim_(\alpha\to(0) )\frac(\arcsin\alpha)(\alpha)=1 \end(equation) \begin(equation) \lim_(\alpha\to(0))\frac(\arctg\alpha)(\alpha)=1 \end(equation)
Eleven examples are solved on this page. Example No. 1 is devoted to the proof of formulas (2)-(4). Examples No. 2, No. 3, No. 4 and No. 5 contain solutions with detailed comments. Examples No. 6-10 contain solutions with virtually no comments, because detailed explanations were given in previous examples. The solution uses some trigonometric formulas that can be found.
I note that the presence trigonometric functions coupled with the uncertainty $\frac (0) (0)$ does not yet mean the mandatory application of the first remarkable limit. Sometimes simple things are enough trigonometric transformations, - 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 obtain:
$$ \lim_(x\to(0))\frac(\frac(\sin(5x))(5x)\cdot(5x))(\frac(\tg(8x))(8x)\cdot(8x )) =\frac(5)(8)\cdot\lim_(x\to(0))\frac(\frac(\sin(5x))(5x))(\frac(\tg(8x))( 8x)) $$
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 in this example there is no point in replacing it, 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: $\lim_(x\to\frac(\pi)(2))\frac(1-\sin(x))(\cos^2x)=\frac(1)(2)$, $\lim_( x\to\frac(2\pi)(3))\frac(\tg(x)+\sqrt(3))(2\cos(x)+1)=-\frac(4)(\sqrt( 3))$.
For those who want to learn how to find limits, in this article we will tell you about it. We won’t delve into the theory; teachers usually give it at lectures. So the “boring theory” should be jotted down in your notebooks. If this is not the case, then you can read textbooks borrowed from the library. educational institution or on other Internet resources.
So, the concept of limit is quite important in the study of higher mathematics, especially when you come across integral calculus and understand the connection between limit and integral. In the current material we will consider simple examples, as well as ways to solve them.
Examples of solutions
| 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)$$ \lim_(x \to \infty) \frac(1)(x) = 0 $$ People often send us these limits with a request to help solve them. We decided to highlight them as a separate example and explain that these limits just need to be remembered, as a rule. If you cannot solve your problem, then send it to us. We will provide detailed solution. You will be able to view the progress of the calculation and gain information. This will help you get your grade from your teacher in a timely manner! |
| Answer |
| $$ \text(a)) \lim \limits_(x \to 0) \frac(1)(x) = \infty \text( b))\lim \limits_(x \to \infty) \frac(1 )(x) = 0 $$ |
What to do with 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 $ 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 now? What should happen in the end? Since this is uncertainty, this is not an answer yet and we continue the calculation. Since we have a polynomial in the numerators, we will factorize it using the formula familiar to everyone from school $$ a^2-b^2=(a-b)(a+b) $$. Do you remember? Great! Now go ahead and use it with the song :) We find that the numerator $ x^2-1=(x-1)(x+1) $ We continue to solve taking into account 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 push 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? What should I do? Don't panic, because the impossible is possible. It is necessary to take out the x in both the numerator and the denominator, and then reduce it. After this, try to calculate the limit. Let's try... $$ \lim \limits_(x \to \infty) \frac(x^2-1)(x+1) =\lim \limits_(x \to \infty) \frac(x^2(1-\frac (1)(x^2)))(x(1+\frac(1)(x))) = $$ $$ = \lim \limits_(x \to \infty) \frac(x(1-\frac(1)(x^2)))((1+\frac(1)(x))) = $$ 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 examples and create an algorithm for solving the limits:
- Substitute point x into the expression following the limit sign. If a certain number or infinity is obtained, then the limit is completely solved. Otherwise, we have uncertainty: “zero divided by zero” or “infinity divided by infinity” and move on to the next steps of the instructions.
- To eliminate the uncertainty of “zero divided by zero,” you need to factor the numerator and denominator. Reduce similar ones. Substitute point x into the expression under the limit sign.
- If the uncertainty is “infinity divided by infinity,” then we take out both the numerator and the denominator x to the greatest degree. We shorten the X's. We substitute the values of x from under the limit into the remaining expression.
In this article, you learned 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'll talk about other types of assignments in future articles, but first you need to learn this lesson in order to move forward. Let's discuss what to do if there are roots, degrees, study infinitesimal equivalent functions, remarkable limits, L'Hopital's rule.
If you can't figure out the limits yourself, don't panic. We are always happy to help!
Function limit- number a will be the limit of some variable value if, in the process of its change, this variable quantity approaches indefinitely 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) functions 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. Here substitute pure 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 the roots of a 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 it specific meaning or a specific area where a function that is limited by a limit falls.
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.
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 several detailed examples of solving 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 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.
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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 in our numerator quadratic equation. 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 to eliminate both types of uncertainty. 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.
Theory of limits- one of the sections mathematical analysis, which some can master, others have difficulty calculating the limits. The question of finding limits is quite general, since there are dozens of techniques solution limits various types. The same limits can be found both using L'Hopital's rule and without it. It happens that scheduling a series of infinitesimal functions allows you to quickly obtain desired result. There are a set of techniques and tricks that allow you to find the limit of a function of any complexity. In this article we will try to understand the main types of limits that are most often encountered in practice. We will not give the theory and definition of the limit here; there are many resources on the Internet where this is discussed. Therefore, let’s get down to practical calculations, this is where your “I don’t know! I can’t! We weren’t taught!”
Calculating limits using the substitution method
Example 1. Find the limit of a function
Lim((x^2-3*x)/(2*x+5),x=3).
Solution: Examples of this kind can be theoretically calculated using the usual substitution
The limit is 18/11.
There is nothing complicated or wise about such limits - we substituted the value, calculated it, and wrote down the limit as an answer. However, based on such limits, everyone is taught that first of all they need to substitute the value into the function. Further, the limits become more complicated, introducing the concept of infinity, uncertainty, and the like.
A limit with uncertainty like infinity divided by infinity. Uncertainty Disclosure Techniques
Example 2. Find the limit of a function
Lim((x^2+2x)/(4x^2+3x-4),x=infinity).
Solution: A limit of the form polynomial divided by a polynomial is given, and the variable tends to infinity 
Simply substituting the value to which the variable should be found to find the limits will not help, we get an uncertainty of the form infinity divided by infinity.
According to the theory of limits, the algorithm for calculating the limit is to find the largest power of “x” in the numerator or denominator. Next, the numerator and denominator are simplified to it and the limit of the function is found 
Since the value tends to zero when the variable approaches infinity, they are neglected, or written into the final expression in the form of zeros 
Immediately from practice, you can get two conclusions that are a hint in the calculations. If a variable tends to infinity and the degree of the numerator is greater than the degree of the denominator, then the limit is equal to infinity. Otherwise, if the polynomial in the denominator is of higher order than in the numerator, the limit is zero.
The limit can be written in formulas like this: 
If we have a function of the form an ordinary field without fractions, then its limit is equal to infinity 
The next type of limits concerns the behavior of functions near zero.
Example 3. Find the limit of a function
Lim((x^2+3x-5)/(x^2+x+2), x=0).
Solution: There is no need to remove the leading factor of the polynomial here. Exactly the opposite, you need to find the smallest power of the numerator and denominator and calculate the limit 
Value x^2; x tend to zero when the variable tends to zero. Therefore, they are neglected, so we get 
that the limit is 2.5.
Now you know how to find the limit of a function of the form, divide a polynomial by a polynomial if the variable tends to infinity or 0. But this is only a small and easy part of the examples. From the following material you will learn how to uncover uncertainties in the limits of a function.
Limit with uncertainty of type 0/0 and methods for its calculation
Everyone immediately remembers the rule that you cannot divide by zero. However, the theory of limits in this context implies infinitesimal functions.
Let's look at a few examples for clarity.
Example 4. Find the limit of a function
Lim((3x^2+10x+7)/(x+1), x=-1).
Solution: When we substitute the value of the variable x = -1 into the denominator, we get zero, and we get the same thing in the numerator. So we have uncertainty of the form 0/0.
Dealing with such uncertainty is simple: you need to factorize the polynomial, or rather, select the factor that turns the function into zero. 
After expansion, the limit of the function can be written as 
That's the whole method for calculating the limit of a function. We do the same if there is a limit of the form polynomial divided by a polynomial.
Example 5. Find the limit of a function
Lim((2x^2-7x+6)/(3x^2-x-10), x=2).
Solution: Direct substitution shows
2*4-7*2+6=0;
3*4-2-10=0
what do we have type 0/0 uncertainty.
Let's divide the polynomials by the factor that introduces the singularity 
There are teachers who teach that polynomials of 2nd order, that is, the “quadratic equations” type, should be solved through the discriminant. But real practice shows that this is longer and more confusing, so get rid of the features within the limits according to the specified algorithm. Thus, we write the function in the form of simple factors and calculate it in the limit 
As you can see, there is nothing complicated in calculating such limits. By the time you study the limits, you know how to divide polynomials, at least according to the program you should have already passed it.
Among the tasks on type 0/0 uncertainty There are some in which you need to use abbreviated multiplication formulas. But if you don’t know them, then by dividing a polynomial by a monomial you can get the desired formula.
Example 6. Find the limit of a function
Lim((x^2-9)/(x-3), x=3).
Solution: We have an uncertainty of type 0/0. In the numerator we use the abbreviated multiplication formula 
and calculate the required limit 
Method for revealing uncertainty by multiplying by its conjugate
The method is applied to the limits in which uncertainty is generated by irrational functions. The numerator or denominator turns to zero at the calculation point and it is not known how to find the boundary.
Example 7. Find the limit of a function
Lim((sqrt(x+2)-sqrt(7x-10))/(3x-6), x=2).
Solution: Let's represent the variable in the limit formula
When substituting, we obtain an uncertainty of type 0/0.
According to the theory of limits, the way to bypass this feature is to multiply the irrational expression by its conjugate. To ensure that the expression does not change, the denominator must be divided by the same value 
Using the difference of squares rule, we simplify the numerator and calculate the limit of the function
We simplify the terms that create the singularity in the limit and perform the substitution
Example 8. Find the limit of a function
Lim((sqrt(x-2)-sqrt(2x-5))/(3-x), x=3).
Solution: Direct substitution shows that the limit has a singularity of the form 0/0. 
To expand, we multiply and divide by the conjugate of the numerator 
We write down the difference of squares
We simplify the terms that introduce the singularity and find the limit of the function
Example 9. Find the limit of a function
Lim((x^2+x-6)/(sqrt(3x-2)-2), x=2).
Solution: Substitute two into the formula 
We get uncertainty 0/0.
The denominator must be multiplied by the conjugate expression, and in the numerator the quadratic equation must be solved or factored, taking into account the singularity. Since it is known that 2 is a root, we find the second root using Vieta’s theorem
Thus, we write the numerator in the form 
and substitute it into the limit
By reducing the difference of squares, we get rid of the singularities in the numerator and denominator
In this way, you can get rid of singularities in many examples, and the application should be noted wherever a given difference of roots turns into zero during substitution. Other types of limits concern exponential functions, infinitesimal functions, logarithms, special limits and other techniques. But you can read about this in the articles listed below about limits.