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Question Number 113742 by bemath last updated on 15/Sep/20

$$\underset{x\to 0}{\lim }\frac{{(1+x)}^{\frac{1}{x}}-e-\frac{ex}{2}}{x^2}=?$$

Answered by bobhans last updated on 15/Sep/20

$$L=\underset{x\to 0}{\lim }\frac{{(1+x)}^{\frac{1}{x}}-e-\frac{ex}{2}}{x^2}$$$$lety={(1+x)}^{\frac{1}{x}}\to \ln y=\frac{1}{x}\ln (1+x)$$$$\ln y=\frac{1}{x}(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...)$$$$\ln y=1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+...$$$$y=e^{(1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+...)}$$$$L=\underset{x\to 0}{\lim }\frac{e^{(1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+...)}-e-\frac{ex}{2}}{x^2}$$$$L=\underset{x\to 0}{\lim }\frac{e(1+(-\frac{x}{2}+\frac{x^2}{3}-...)+\frac{1}{2!}{(-\frac{x}{2}+\frac{x^2}{3}-...)}^2+...)}{x^2}$$$$L=\underset{x\to 0}{\lim }\frac{(e+\frac{ex}{2}+\frac{11{ex}^2}{24}+...)-e-\frac{ex}{2}}{x^2}$$$$L=\underset{x\to 0}{\lim }\frac{\frac{11{ex}^2}{24}}{x^2}=\frac{11e}{24}$$

Answered by mathmax by abdo last updated on 16/Sep/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{{(1+\mathrm{x})}^{\frac{1}{\mathrm{x}}}-\mathrm{e}-\frac{\mathrm{ex}}{2}}{{\mathrm{x}}^2}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{e}}^{\frac{1}{\mathrm{x}}\ln (1+\mathrm{x})}-\mathrm{e}-\frac{\mathrm{ex}}{2}}{{\mathrm{x}}^2}$$$$\mathrm{we}\mathrm{have}\frac{\mathrm{d}}{\mathrm{dx}}\ln (1+\mathrm{x})=\frac{1}{1+\mathrm{x}}=1-\mathrm{x}+{\mathrm{x}}^2+\mathrm{o}\left({\mathrm{x}}^3\right)\Rightarrow$$$$\ln (1+\mathrm{x})=\mathrm{x}-\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^3}{3}+\mathrm{o}\left({\mathrm{x}}^4\right)\Rightarrow \frac{\ln (1+\mathrm{x})}{\mathrm{x}}=1-\frac{\mathrm{x}}{2}+\frac{{\mathrm{x}}^2}{3}+\mathrm{o}\left({\mathrm{x}}^3\right)\Rightarrow$$$${\mathrm{e}}^{\frac{1}{\mathrm{x}}\ln (1+\mathrm{x})}=\mathrm{e}{\mathrm{e}}^{-\frac{\mathrm{x}}{2}+\frac{{\mathrm{x}}^2}{3}+\mathrm{o}\left({\mathrm{x}}^3\right)}\sim \mathrm{e}(1-\frac{\mathrm{x}}{2}+\frac{{\mathrm{x}}^2}{3})=\mathrm{e}-\frac{\mathrm{ex}}{2}+\frac{{\mathrm{ex}}^2}{3}\Rightarrow$$$${\mathrm{e}}^{\frac{1}{\mathrm{x}}\ln (1+\mathrm{x})}-\mathrm{e}+\frac{\mathrm{ex}}{2}\sim \frac{{\mathrm{ex}}^2}{3}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)\sim \frac{\mathrm{e}}{3}\Rightarrow {\lim }_{\mathrm{x}\to 0}\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{e}}{3}$$$$\mathrm{i}\mathrm{think}\mathrm{the}\mathrm{Q}\mathrm{is}{\lim }_{\mathrm{x}\to 0}\frac{{(1+\mathrm{x})}^{\frac{1}{\mathrm{x}}}-\mathrm{e}+\frac{\mathrm{ex}}{2}}{{\mathrm{x}}^2}...!$$

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