The limit of a function, computing limits.

Definition 1. A number $a$ is called the limit of the function $f(x)$ at the point $x_0$ (or as $x$ approaches $x_0$) if for every number $\epsilon >0$, there exists a number $\delta >0$ such that for all $x$ satisfying the condition $0<|x-x_0|<\delta$, the inequality $|f(x)-a|<\epsilon$ holds.

The number $a$ is the limit of the function $f(x)$ at the point $x_0$, denoted as $\underset{x\to x_0}{lim}f(x)=a$, if for every $\epsilon >0$, there exists $\delta >0$ such that for all $x:0<|x-x_0|<\delta \Rightarrow |f(x)-a|<\epsilon$.

The number $a$ is not the limit of the function $f(x)$ at the point $x_0$, denoted as $\underset{x\to x_0}{lim}f(x)\ne a$, if there exists $\epsilon >0$ such that for every $\delta >0$, there exists $x:0<|x-x_0|<\delta \Rightarrow |f(x)-a|\ge \epsilon$.

Properties of the limit:

1) If $f(x)$ and $g(x)$ have limits at the point $x_0$, then the functions $f(x)\pm g(x)$ and $f(x)g(x)$ also have limits at the point $x_0$, and

$$\underset{x\to x_0}{lim}(f(x)\pm g(x))=\underset{x\to x_0}{lim}f(x)\pm \underset{x\to x_0}{lim}g(x);$$

$$\underset{x\to x_0}{lim}(f(x)g(x))=(\underset{x\to x_0}{lim}f(x))(\underset{x\to x_0}{lim}g(x))$$

2) For any number $C,\underset{x\to x_0}{lim}(Cf(x))=C\underset{x\to x_0}{lim}f(x)$

3) If the functions $f(x)$ and $g(x)$ have limits at the point $x_0$ and $\underset{x\to x_0}{lim}g(x)\ne 0$, then the function $\frac{f(x)}{g(x)}$ also has a limit at the point $x_0$, and

$$\underset{x\to x_0}{lim}\frac{f(x)}{g(x)}=\frac{\underset{x\to x_0}{lim}f(x)}{\underset{x\to x_0}{lim}g(x)}.$$

4) Let $\underset{x\to x_0}{lim}f(x)=a$ ($f(x)\ne a$ for $x\ne x_0$) and $\underset{y\to a}{lim}g(y)$ exist; then at the point $x_0$, the limit of the composition $g(f(x))$ exists, and $\underset{x\to x_0}{lim}g(f(x))=\underset{y\to a}{lim}g(y).$

If the difference $f(x)-g(x)$ represents an indeterminate form of $\mathrm{\infty }-\mathrm{\infty }$, or the quotient $\frac{f(x)}{g(x)}$ represents an indeterminate form of $\frac{\mathrm{\infty }}{\mathrm{\infty }}$ or $\frac{0}{0}$, then the computation of limits is called "indeterminacy resolution."

Examples.

1) $$\underset{x\to 1}{lim}\frac{x^2-4}{x^2-x-2}=\frac{\underset{x\to 1}{lim}(x^2-4)}{\underset{x\to 1}{lim}(x^2-x-2)}=\frac{1-4}{1-1-2}=\frac{3}{2}.$$

2) $\underset{x\to 2}{lim}\frac{x^2-4}{x^2-x-2}$

$$\underset{x\to 2}{lim}(x^2-4)=4-4=0;\underset{x\to 2}{lim}(x^2-x-2)=4-2-2=0.$$ Таким образом, имеем неопределенность вида $\frac{0}{0}.$

$$\underset{x\to 2}{lim}\frac{x^2-4}{x^2-x-2}=\left[\frac{0}{0}\right]=\underset{x\to 2}{lim}\frac{(x-2)(x+2)}{(x-2)(x+1)}=\underset{x\to 2}{lim}\frac{x+2}{x+1}=\frac{4}{3}.$$

3) $$\underset{x\to \mathrm{\infty }}{lim}\frac{x^2-4}{x^2-x-2}=[\underset{x\to \mathrm{\infty }}{lim}(x^2-4)=\underset{x\to \mathrm{\infty }}{lim}(x^2-x-2)=\mathrm{\infty }].$$

Следовательно $$\underset{x\to \mathrm{\infty }}{lim}\frac{x^2-4}{x^2-x-2}=\left[\frac{\mathrm{\infty }}{\mathrm{\infty }}\right]=\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{x^2}{x^2}-\frac{4}{x^2}}{\frac{x^2}{x^2}-\frac{x}{x^2}-\frac{2}{x^2}}=1.$$

4) $$\underset{x\to 1}{lim}\frac{x^2+4x-5}{x^2-1}=\left[\frac{0}{0}\right]=\underset{x\to 1}{lim}\frac{(x-1)(x+5)}{(x-1)(x+1)}=\underset{x\to 1}{lim}\frac{x+5}{x+1}=\frac{6}{2}=3.$$

5) $$\underset{x\to \mathrm{\infty }}{lim}\frac{x^2+4x-5}{x^2-1}=\left[\frac{\mathrm{\infty }}{\mathrm{\infty }}\right]=\underset{x\to \mathrm{\infty }}{lim}\frac{1-\frac{4}{x}-\frac{5}{x^2}}{1-\frac{1}{x^2}}=1.$$

6) $$\underset{x\to -1}{lim}\frac{x^2+4x-5}{x^2-1}=\frac{1-4-5}{0}=\mathrm{\infty }.$$

7) $$\underset{x\to \mathrm{\infty }}{lim}(\frac{x^3+3x^2}{x^2+1}-x)=[\mathrm{\infty }-\mathrm{\infty }]=\underset{x\to \mathrm{\infty }}{lim}\frac{x^3+3x^2-x^3-x}{x^2+1}=$$ $$=\underset{x\to \mathrm{\infty }}{lim}\frac{3x^2-x}{x^2+1}=\underset{x\to \mathrm{\infty }}{lim}\frac{3-\frac{x}{x^2}}{1+\frac{1}{x^2}}=3.$$

8) $$\underset{x\to 6}{lim}\frac{\sqrt{x-2}-2}{x-6}=\left[\frac{0}{0}\right]=\underset{x\to 6}{lim}\frac{(\sqrt{x-2}-2)(\sqrt{x-2}+2)}{(x-6)(\sqrt{x-2}+2)}=$$ $$=\underset{x\to 6}{lim}\frac{x-2-4}{(x-6)(\sqrt{x-2}+2)}=\underset{x\to 6}{lim}\frac{1}{\sqrt{x-2}+2}=\frac{1}{4}.$$

9) $$\underset{x\to \mathrm{\infty }}{lim}\frac{5-x^6}{\sqrt{x^{12}+5x^5-1}}=\left[\frac{\mathrm{\infty }}{\mathrm{\infty }}\right]=\underset{x\to \mathrm{\infty }}{lim}\frac{5-\frac{1}{x^6}}{\sqrt{1+\frac{5}{x^7}-\frac{1}{x^{12}}}}=5.$$

10) $$\underset{x\to \mathrm{\infty }}{lim}(\sqrt{x^4+2x^2-1}-\sqrt{x^4-2x^2-1})=[\mathrm{\infty }-\mathrm{\infty }]=$$

$$=\underset{x\to \mathrm{\infty }}{lim}\frac{(\sqrt{x^4+2x^2-1}-\sqrt{x^4-2x^2-1})(\sqrt{x^4+2x^2-1}+\sqrt{x^4-2x^2-1})}{\sqrt{x^4+2x^2-1}+\sqrt{x^4-2x^2-1}}=$$

$$=\underset{x\to \mathrm{\infty }}{lim}\frac{x^4+2x^2-1-x^4+2x^2+1}{\sqrt{x^4+2x^2-1}+\sqrt{x^4-2x^2-1}}=$$

$$=\left[\frac{\mathrm{\infty }}{\mathrm{\infty }}\right]=\underset{x\to \mathrm{\infty }}{lim}\frac{4x^2}{\sqrt{x^4+2x^2-1}+\sqrt{x^4-2x^2-1}}=\underset{x\to \mathrm{\infty }}{lim}\frac{4}{\sqrt{1+\frac{2}{x^2}-\frac{1}{x^4}}+\sqrt{1-\frac{2}{x^2}-\frac{1}{x^4}}}=2.$$

Some remarkable limits.

The computation of limits in many cases is carried out using two important formulas:

$$\underset{x\to 0}{lim}\frac{\sin x}{x}=1;\underset{x\to 0}{lim}(1+x{)}^{\frac{1}{x}=e}.$$

The following formulas are often used, which are consequences of the above-mentioned formulas.

$$\underset{x\to \mathrm{\infty }}{lim}{(1+\frac{1}{x})}^x=e.$$

$$\underset{x\to 0}{lim}\frac{{\log }_a(1+x)}{x}=\frac{1}{\ln a},a>0,a\ne 1.$$

A special case when $a=e:$ $$\underset{x\to 0}{lim}\frac{\ln (1+x)}{x}=1.$$

$$\underset{x\to 0}{lim}\frac{a^x-1}{x}=\ln a,a>0.$$

In the special case when $a=e:$ $$\underset{x\to 0}{lim}\frac{e^x-1}{1}=1.$$

Thus, we can write down the following equivalent functions:

$$\sin x\sim tgx\sim arctgx\sim arcsinx\sim x(x\to 0);$$

$$1-\cos x\sim \frac{x^2}{2}(x\to 0);$$

$$e^x-1\sim \ln (1+x)\sim x(x\to 0);$$

$$a^x-1\sim x\ln a(x\to 0);$$

$${\log }_a(1+x)\sim x{\log }_{a}e(x\to 0).$$

Examples.

$$1)\underset{x\to 0}{lim}\frac{\sin ax}{x}=\underset{x\to 0}{lim}\frac{a\sin ax}{ax}=a\underset{x\to 0}{lim}\frac{\sin ax}{ax}=a.$$

Another method:

$$\underset{x\to 0}{lim}\frac{\sin ax}{x}=[\sin ax\sim ax(x\to 0)]=\underset{x\to 0}{lim}\frac{ax}{x}=a.$$

$$2)\underset{x\to 0}{lim}\frac{arctgx}{x}=[y=arctgx;x=tgy]=\underset{y\to 0}{lim}\frac{y}{tgy}=\underset{y\to 0}{lim}\frac{y}{\frac{\sin y}{\cos y}}=\underset{y\to 0}{lim}\frac{\cos y}{\frac{\sin y}{y}}=1.$$

$$3)\underset{x\to 0}{lim}\frac{tg4x}{\sin x}=\underset{x\to 0}{lim}\left(\frac{\sin 4x}{4x}\frac{4x}{\cos 4x\sin x}\right)=4.$$

$$4)\underset{x\to \mathrm{\infty }}{lim}{\left(\frac{x}{2x+1}\right)}^x=\underset{x\to \mathrm{\infty }}{lim}{\left({\left(\frac{2x}{2x+1}\right)}^{2x}\right)}^{\frac{1}{2}}\frac{1}{2^x}=\underset{x\to 0}{lim}{\left({\left(\frac{1}{1+\frac{1}{2x}}\right)}^{2x}\right)}^{\frac{1}{2}}\frac{1}{2^x}=$$ $$e^{-1/2}\underset{x\to \mathrm{\infty }}{lim}\frac{1}{2^x}=0.$$

$$5)\underset{x\to 0}{lim}\frac{e^{x^2}-\cos x}{{\sin }^{2}x}=\underset{x\to 0}{lim}\frac{e^{x^2}-1+1-\cos x}{x^2}=\underset{x\to 0}{lim}\frac{e^{x^2}-1}{x^2}+\underset{x\to 0}{lim}\frac{1-\cos x}{x^2}=$$ $$=[e^{x^2}-1\sim -x^2,(x\to 0);1-\cos x\sim \frac{x^2}{2},x\to 0]=$$

$$=\underset{x\to 0}{lim}\frac{-x^2}{x^2}+\underset{x\to 0}{lim}\frac{x^2/2}{x^2}=-1+\frac{1}{2}=\frac{1}{2}.$$

$$6)\underset{x\to 0}{lim}(\ln (e+tg\alpha x){)}^{1/\sin \beta x}=\underset{x\to 0}{lim}{(\ln (e(1+\frac{tg\alpha x}{e}))}^{\frac{1}{\sin \beta x}}=$$

$$=\underset{x\to 0}{lim}{(\ln e+\ln (1+\frac{tg\alpha x}{e}))}^{\frac{1}{\sin \beta x}}=\underset{x\to 0}{lim}{(1+\ln (1+\frac{tg\alpha x}{e}))}^{\frac{1}{\sin \beta x}}.$$

As $x\to 0,$ $\ln (1+\frac{tg\alpha x}{e})\sim \frac{tg\alpha x}{e}\to 0.$ Therefore,

$$\underset{x\to 0}{lim}{(1+\ln (1+\frac{tg\alpha x}{e}))}^{\frac{1}{\sin \beta x}}=$$

$$=\underset{x\to 0}{lim}{(1+\ln (1+\frac{tg\alpha x}{e}))}^{\frac{1}{\ln (1+\frac{tg\alpha x}{e})}\ln (1+\frac{tg\alpha x}{e})\frac{1}{\sin \beta x}}=\underset{x\to 0}{lim}{e}^{\frac{\ln (1+\frac{tg\alpha x}{e})}{\sin \beta x}}=$$

$$=e^{\underset{x\to 0}{lim}\frac{\ln (1+\frac{tg\alpha x}{e})}{\sin \beta x}}=[\ln (1+\frac{tg\alpha x}{e})\sim \frac{\alpha x}{e},(x\to 0);\sin \beta x\sim \beta x,(x\to 0)]=$$ $$e^{\underset{x\to 0}{lim}\frac{\alpha x}{e\beta x}}=e^{\frac{\alpha }{e\beta }}$$

$$7)\underset{x\to 0}{lim}ct{g}^{2}x(3^{\cos x}-3)=\underset{x\to 0}{lim}\frac{3^{\cos x}-3}{x^2}=\underset{x\to 0}{lim}\frac{3(3^{\cos x-1}-1)}{x^2}=\underset{x\to 0}{lim}\frac{3(\cos x-1)\ln 3}{x^2}=$$ $$\underset{x\to 0}{lim}\frac{3\ln 3(-{\sin }^2\frac{x}{2})}{x^2}=-6\ln 3\cdot \frac{1}{4}=-\frac{3}{2}\ln 3$$

Tags: Properties of the limit, computing limits, limit, limit of function, calculus, mathematical amalysis