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 $\varepsilon > 0$, there exists a number $\delta > 0$ such that for all $x$ satisfying the condition $0 < |x - x_0| < \delta$, the inequality $|f(x) - a| < \varepsilon$ holds.

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

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

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

$$\lim\limits_{x\rightarrow x_0}(f(x)\pm g(x))=\lim\limits_{x\rightarrow x_0}f(x)\pm\lim\limits_{x\rightarrow x_0}g(x);$$

$$\lim\limits_{x\rightarrow x_0}(f(x)g(x))=(\lim\limits_{x\rightarrow x_0}f(x))(\lim\limits_{x\rightarrow x_0}g(x))$$

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

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

$$\lim\limits_{x\rightarrow x_0}\frac{f(x)}{g(x)}=\frac{\lim\limits_{x\rightarrow x_0}f(x)}{\lim\limits_{x\rightarrow x_0}g(x)}.$$

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

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

Examples.

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

2) $\lim\limits_{x\rightarrow 2}\frac{x^2-4}{x^2-x-2}$

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

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

3) $$\lim\limits_{x\rightarrow \infty}\frac{x^2-4}{x^2-x-2}=[\lim\limits_{x\rightarrow \infty}(x^2-4)=\lim\limits_{x\rightarrow \infty}(x^2-x-2)=\infty].$$

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

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

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

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

7) $$\lim\limits_{x\rightarrow \infty}\left(\frac{x^3+3x^2}{x^2+1}-x\right)=[\infty-\infty]=\lim\limits_{x\rightarrow \infty}\frac{x^3+3x^2-x^3-x}{x^2+1}=$$ $$=\lim\limits_{x\rightarrow \infty}\frac{3x^2-x}{x^2+1}=\lim\limits_{x\rightarrow \infty}\frac{3-\frac{x}{x^2}}{1+\frac{1}{x^2}}=3.$$

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

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

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

$$=\lim\limits_{x\rightarrow\infty}\frac{(\sqrt{x^4+2x^2-1}-\sqrt{x^4-2x^2-1})(\sqrt{x^4+2x^2-1}+\sqrt{x^4-2x^2-1})}{\sqrt{x^4+2x^2-1}+\sqrt{x^4-2x^2-1}}=$$

$$=\lim\limits_{x\rightarrow \infty}\frac{x^4+2x^2-1-x^4+2x^2+1}{\sqrt{x^4+2x^2-1}+\sqrt{x^4-2x^2-1}}=$$

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

Some remarkable limits.

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

$$\lim\limits_{x\rightarrow 0}\frac{\sin x}{x}=1; \,\,\,\,\,\,\, \lim\limits_{x\rightarrow 0}(1+x)^{\frac{1}{x}=e}.$$

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

$$\lim\limits_{x\rightarrow \infty}\left(1+\frac{1}{x}\right)^x=e.$$

$$\lim\limits_{x\rightarrow 0}\frac{\log_a(1+x)}{x}=\frac{1}{\ln a},\,\,\, a>0, a\neq 1.$$

Частный случай при $a=e:$ $$\lim\limits_{x\rightarrow 0}\frac{\ln{(1+x)}}{x}=1.$$

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

В частный случае, когда $a=e:$ $$\lim\limits_{x\rightarrow 0}\frac{e^x-1}{1}=1.$$

Thus, we can write down the following equivalent functions:

$$\sin x\sim tg x\sim arctg x\sim arcsin x\sim x \qquad (x\rightarrow 0);$$

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

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

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

$$\log_a(1+x)\sim x \log_a e \qquad (x\rightarrow 0).$$

Examples.

$$1)\lim\limits_{x\rightarrow 0}\frac{\sin ax}{x}=\lim\limits_{x\rightarrow 0}\frac{a\sin ax}{ax}=a\lim\limits_{x\rightarrow 0}\frac{\sin ax}{ax}=a.$$

Another method:

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

$$2)\lim\limits_{x\rightarrow 0}\frac{arctg x}{x}=[y=arctg x;\,\, x=tg y]=\lim\limits_{y\rightarrow 0}\frac{y}{tg y}=\lim\limits_{y\rightarrow 0}\frac{y}{\frac{\sin y}{\cos y}}=\lim\limits_{y\rightarrow 0}\frac{\cos y}{\frac{\sin y}{y}}=1.$$

$$3)\lim\limits_{x\rightarrow 0}\frac{tg 4x}{\sin x}=\lim\limits_{x\rightarrow 0}\left(\frac{\sin 4x}{4x}\frac{4x}{\cos 4x \sin x}\right)=4.$$

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

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

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

$$6)\lim\limits_{x\rightarrow 0}(\ln(e+tg \alpha x))^{1/\sin \beta x}=\lim\limits_{x\rightarrow 0}\left(\ln(e\left(1+\frac{tg\alpha x}{e}\right)\right)^{\frac{1}{\sin\beta x}}=$$

$$=\lim\limits_{x\rightarrow 0}\left(\ln e+\ln\left(1+\frac{tg\alpha x}{e}\right)\right)^{\frac{1}{\sin\beta x}}=\lim\limits_{x\rightarrow 0}\left(1+\ln\left(1+\frac{tg\alpha x}{e}\right)\right)^{\frac{1}{\sin\beta x}}.$$

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

$$\lim\limits_{x\rightarrow 0}\left(1+\ln\left(1+\frac{tg\alpha x}{e}\right)\right)^{\frac{1}{\sin\beta x}}=$$

$$=\lim\limits_{x\rightarrow 0}\left(1+\ln\left(1+\frac{tg\alpha x}{e}\right)\right)^{\frac{1}{\ln\left(1+\frac{tg\alpha x}{e}\right)}\ln\left(1+\frac{tg\alpha x}{e}\right)\frac{1}{\sin\beta x}}=\lim\limits_{x\rightarrow 0}e^{\frac{\ln\left(1+\frac{tg\alpha x}{e}\right)}{\sin\beta x}}=$$

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

$$7)\lim\limits_{x\rightarrow 0}ctg^2 x(3^{\cos x}-3)=\lim\limits_{x\rightarrow 0}\frac{3^{\cos x}-3}{x^2}=\lim\limits_{x\rightarrow 0}\frac{3(3^{\cos x-1}-1)}{x^2}=\lim\limits_{x\rightarrow 0}\frac{3(\cos x-1)\ln 3}{x^2}=$$ $$\lim\limits_{x\rightarrow 0}\frac{3\ln 3(-\sin^2\frac{x}{2})}{x^2}=-6\ln 3\cdot\frac{1}{4}=-\frac{3}{2}\ln 3$$