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Question Number 107745 by ajfour last updated on 12/Aug/20

$$\left(i\right)L=\underset{x\to 0}{\lim }{\left[\frac{{(1+x)}^{1/x}}{e}\right]}^{1/x}=?$$$$\left(ii\right)L=\underset{x\to \mathrm{\infty }}{\lim }[\frac{x}{e}-x{\left(\frac{x}{x+1}\right)}^x]=?$$

Commented by ajfour last updated on 12/Aug/20

$$MJSSir,pleasehelp..$$

Commented by bemath last updated on 13/Aug/20

Commented by bobhans last updated on 13/Aug/20

$$\mathrm{wow}..==\mathrm{great}\mathrm{and}\mathrm{nice}$$

Answered by Dwaipayan Shikari last updated on 13/Aug/20

$$\underset{x\to 0}{\lim }{\left(\frac{{(1+x)}^{\frac{1}{x}}}{e}\right)}^{\frac{1}{x}}=L$$$$\frac{1}{x}log{(1+x)}^{\frac{1}{x}}-\frac{1}{x}loge=logL$$$$\frac{log(1+x)}{x^2}-\frac{1}{x}=logL$$$$\frac{log(1+x^2)-x}{x^2}=logL$$$$\frac{\frac{1}{1+x}-1}{2x}=logL$$$$\frac{\frac{-1}{{(1+x)}^2}}{2}=logL$$$$-\frac{1}{2}=logL$$$$L=\frac{1}{\sqrt{e}}$$

Commented by ajfour last updated on 12/Aug/20

$$Great!thanksalot.$$

Commented by Dwaipayan Shikari last updated on 12/Aug/20

$$Anotherway$$$$\frac{log(1+x)}{x^2}-1=(1-\frac{1}{2}+\frac{x}{3}-....)-1=-\frac{1}{2}$$$$L=e^{\frac{-1}{2}}=\frac{1}{\sqrt{e}}$$

Commented by bemath last updated on 13/Aug/20

$$howdoyouget\frac{1}{x}\log {(1+x)}^{\frac{1}{x}}-\frac{1}{x}\log e^x$$$$itwrong=$$$$theoriginallimit$$$$\underset{x\to 0}{\lim }{\left(\frac{{(1+x)}^{\frac{1}{x}}}{e}\right)}^{\frac{1}{x}}not$$$$\underset{x\to 0}{\lim }{\left(\frac{{(1+x)}^{\frac{1}{x}}}{e^x}\right)}^{\frac{1}{x}}.$$

Answered by mathmax by abdo last updated on 12/Aug/20

$$\mathrm{f}\left(\mathrm{x}\right)={\left\{\frac{{(1+\mathrm{x})}^{\frac{1}{\mathrm{x}}}}{\mathrm{e}}\right\}}^{\frac{1}{\mathrm{x}}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{{(1+\mathrm{x})}^{\frac{1}{{\mathrm{x}}^2}}}{{\mathrm{e}}^{\frac{1}{\mathrm{x}}}}={\mathrm{e}}^{-\frac{1}{\mathrm{x}}}{(1+\mathrm{x})}^{\frac{1}{{\mathrm{x}}^2}}$$$$={\mathrm{e}}^{-\frac{1}{\mathrm{x}}}.{\mathrm{e}}^{\frac{1}{{\mathrm{x}}^2}\ln (1+\mathrm{x})}\mathrm{we}\mathrm{have}{\ln }^{{}^{\prime }}(1+\mathrm{x})=\frac{1}{1+\mathrm{x}}=1-\mathrm{x}+\mathrm{o}\left(\mathrm{x}\right)\Rightarrow$$$$\ln (1+\mathrm{x})=\mathrm{x}-\frac{{\mathrm{x}}^2}{2}+\mathrm{o}\left({\mathrm{x}}^2\right)\Rightarrow \frac{1}{{\mathrm{x}}^2}\ln (1+\mathrm{x})=\frac{1}{\mathrm{x}}-\frac{1}{2}+\mathrm{o}\left(1\right)\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{-\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{x}}-\frac{1}{2}+\mathrm{o}\left(1\right)}={\mathrm{e}}^{-\frac{1}{2}+\mathrm{o}\left(1\right)}\Rightarrow {\lim }_{\mathrm{x}\to 0}\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\sqrt{\mathrm{e}}}$$

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