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Question Number 87168 by mathocean1 last updated on 03/Apr/20

$$Calculatetheselimits:$$$$Pleasesirsdetail$$

Commented by john santu last updated on 03/Apr/20

$$\mathrm{a})\mathrm{change}\mathrm{x}\Rightarrow -\mathrm{t}$$$$\underset{\mathrm{t}\to \mathrm{\infty }}{\lim }\frac{-3\mathrm{t}-1-\cos \mathrm{t}}{-\mathrm{t}}=3$$$$\mathrm{c})\underset{x\to 4}{\lim }\frac{\frac{1}{\sqrt{2\mathrm{x}+1}}}{1-\frac{3}{2\sqrt{3\mathrm{x}+4}}}=\frac{\frac{1}{3}}{1-\frac{3}{8}}=\frac{\frac{1}{3}}{\frac{5}{8}}=\frac{8}{15}$$$$\mathrm{d})\underset{x\to \mathrm{\infty }}{\lim }\frac{\mathrm{x}\sqrt{3+\frac{1}{{\mathrm{x}}^2}}-2\mathrm{x}}{\mathrm{x}}=\sqrt{3}-2$$

Commented by mathocean1 last updated on 03/Apr/20

Commented by john santu last updated on 03/Apr/20

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\mathrm{or}\underset{x\to \mathrm{\infty }}{\lim }$$

Commented by mathocean1 last updated on 03/Apr/20

$$li\underset{x\to \mathrm{\infty }}{m}$$

Commented by TANMAY PANACEA. last updated on 03/Apr/20

$$c)\underset{x\to 4}{\lim }\frac{\sqrt{2x+1}-3}{x-\sqrt{3x+4}}$$$$=\underset{x\to 4}{\lim }\frac{2x+1-9}{x^2-3x-4}\times \frac{x+\sqrt{3x+4}}{\sqrt{2x+1}+3}$$$$=\underset{x\to 4}{\lim }\frac{2(x-4)}{(x-4)(x+1)}\times \frac{x+\sqrt{3x+4}}{\sqrt{2x+1}+3}$$$$=\frac{2}{5}\times \frac{4+4}{3+3}=\frac{2}{5}\times \frac{4}{3}=\frac{8}{15}$$

Commented by TANMAY PANACEA. last updated on 03/Apr/20

$$d)\underset{x\to \mathrm{\infty }}{\lim }\frac{\sqrt{3x^2+1}-2x}{x}$$$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{x\sqrt{3+\frac{1}{x^2}}-2x}{x}$$$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{\sqrt{3+\frac{1}{x^2}}-2}{1}=\sqrt{3}-2$$

Commented by mathmax by abdo last updated on 03/Apr/20

$$a)wehave{lim}_{x\to -\mathrm{\infty }}\frac{3x+1-cosx}{x}$$$$={lim}_{x\to -\mathrm{\infty }}\frac{x(3+\frac{1}{x}-\frac{cosx}{x})}{x}={lim}_{x\to -\mathrm{\infty }}3+\frac{1}{x}-\frac{cosx}{x}=3$$

Commented by mathmax by abdo last updated on 03/Apr/20

$$c){lim}_{x\to 4}\frac{\sqrt{2x+1}-3}{x-\sqrt{3x+4}}={lim}_{x\to 4}\frac{2x+1-9}{\sqrt{2x+1}+3}\times \frac{x+\sqrt{3x+4}}{x^2-3x-4}$$$$={lim}_{x\to 4}\frac{x+\sqrt{3x+4}}{\sqrt{2x+1}+3}\times \frac{2x-8}{(x^2-3x-4)}=\frac{4}{3}{lim}_{x\to 4}\frac{2(x-4)}{(x-4)(x+1)}$$$$=\frac{4}{3}{lim}_{x\to 4}\frac{2}{x+1}=\frac{4}{3}\times \frac{2}{5}=\frac{8}{15}$$

Commented by mathmax by abdo last updated on 03/Apr/20

$$d)f\left(x\right)=\frac{\sqrt{3x^2+1}-2x}{x}wehaveforx>0$$$$\sqrt{3x^2+1}=x\sqrt{3}\sqrt{1+\frac{1}{3x^2}}\sim x\sqrt{3}\{1+\frac{1}{6x^2}\}\Rightarrow f\left(x\right)\sim \frac{x\sqrt{3}+\frac{\sqrt{3}}{6x}-2x}{x}$$$$=\frac{\sqrt{3}+\frac{\sqrt{3}}{6x^2}-2}{1}\to \sqrt{3}-2\Rightarrow {lim}_{x\to +\mathrm{\infty }}f\left(x\right)=\sqrt{3}-2$$

Commented by mathmax by abdo last updated on 03/Apr/20

$$b)letf\left(x\right)=\frac{x^2+1}{x\left[x\right]}changementx=-tgive$$$$f\left(x\right)=\frac{t^2+1}{-t[-t]}(x\to -\mathrm{\infty }\Rightarrow t\to +\mathrm{\infty })andwehave$$$$[-t]=\alpha -\left[t\right]with\alpha =0or-1\Rightarrow [-t]\sim -\left[t\right]\sim -t\Rightarrow$$$${lim}_{x\to -\mathrm{\infty }}f\left(x\right)={lim}_{t\to +\mathrm{\infty }}-\frac{t^2+1}{t(-t)}={lim}_{t\to +\mathrm{\infty }}\frac{t^2+1}{t^2}=1$$$$$$

Commented by john santu last updated on 03/Apr/20

$$\mathrm{b})\mathrm{what}\mathrm{define}\mathrm{E}\left(\mathrm{x}\right)?$$

Commented by john santu last updated on 03/Apr/20

$$\mathrm{why}\mathrm{E}\left(\mathrm{x}\right)=\left[\mathrm{x}\right]?$$$$\mathrm{E}\left(\mathrm{x}\right)=\mu ?$$

Commented by mathmax by abdo last updated on 03/Apr/20

$$E\left(x\right)=xat\mathrm{\infty }$$

Commented by mathmax by abdo last updated on 03/Apr/20

$$x=\left[x\right]+rwith0⩽r<1\Rightarrow \frac{x}{\left[x\right]}=1+\frac{r}{\left[x\right]}\Rightarrow \frac{x}{\left[x\right]}\sim 1(x\to +\mathrm{\infty })$$

Commented by mathocean1 last updated on 03/Apr/20

$$\mathrm{E}\left(\mathrm{x}\right)⩽\mathrm{x}⩽\mathrm{E}\left(\mathrm{x}\right)+1$$

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