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Question Number 92577 by jagoll last updated on 08/May/20

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\ln \mathrm{x}}{\mathrm{x}}$$

Commented by Tony Lin last updated on 08/May/20

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{lnx}{x}$$$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{\frac{1}{x}}{1}\left(Hospitaltheorem\right)$$$$=0$$

Commented by jagoll last updated on 08/May/20

$$\mathrm{thank}\mathrm{you}.$$$$1\mathrm{way}\mathrm{via}{\mathrm{L}}^{\prime }\mathrm{Hopital}$$$$1\mathrm{way}\mathrm{via}\mathrm{squeeze}\mathrm{theorem}$$

Commented by turbo msup by abdo last updated on 08/May/20

$$letx=\frac{1}{t}x\to +\mathrm{\infty }\Rightarrow t\to 0$$$${lim}_{x\to +\mathrm{\infty }}\frac{lnx}{x}={lim}_{t\to 0^{+}}-tln\left(t\right)=0$$

Answered by john santu last updated on 08/May/20

$$\Rightarrow \ln \left(\mathrm{x}\right)={\int }_1^x\frac{1}{\mathrm{t}}\mathrm{dt}⩽\int \underset{1}{\stackrel{\mathrm{x}}{}}1\mathrm{dt}$$$$\ln \left(\mathrm{x}\right)⩽\mathrm{x},\mathrm{\forall }\mathrm{x}⩾1$$$$\ln \left(\sqrt{\mathrm{x}}\right)⩽\sqrt{\mathrm{x}}$$$$\frac{\ln \left(\mathrm{x}\right)}{2}⩽\sqrt{\mathrm{x}}\Rightarrow \ln \left(\mathrm{x}\right)⩽2\sqrt{\mathrm{x}}$$$$\frac{\ln \left(\mathrm{x}\right)}{\mathrm{x}}⩽\frac{2\sqrt{\mathrm{x}}}{\mathrm{x}}=\frac{2}{\sqrt{\mathrm{x}}}$$$$\mathrm{by}\mathrm{squeeze}\mathrm{theorem}$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\ln \left(\mathrm{x}\right)}{\mathrm{x}}⩽\underset{x\to \mathrm{\infty }}{\lim }\frac{2}{\sqrt{\mathrm{x}}}$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\ln \left(\mathrm{x}\right)}{\mathrm{x}}=0$$

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