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Question Number 172910 by DAVONG last updated on 03/Jul/22

$$\underset{x\to 1}{\lim }\frac{\mathrm{e}-{\mathrm{x}}^{\frac{1}{\mathrm{x}-1}}}{\mathrm{x}-1}=?$$

Answered by FongXD last updated on 03/Jul/22

$$\mathrm{L}=\underset{\mathrm{x}\to 1}{\lim }\frac{\mathrm{e}-{\mathrm{e}}^{\frac{1}{\mathrm{x}-1}\mathrm{lnx}}}{\mathrm{x}-1}=-\underset{\mathrm{x}\to 1}{\lim }\frac{\mathrm{e}({\mathrm{e}}^{\frac{\mathrm{lnx}}{\mathrm{x}-1}-1}-1)}{\frac{\mathrm{lnx}}{\mathrm{x}-1}-1}\times \frac{\frac{\mathrm{lnx}}{\mathrm{x}-1}-1}{\mathrm{x}-1}$$$$\mathrm{L}=-\mathrm{e}\underset{\mathrm{x}\to 1}{\lim }\frac{\mathrm{lnx}-\mathrm{x}+1}{{(\mathrm{x}-1)}^2}$$$$\mathrm{put}\mathrm{x}={\mathrm{e}}^{\mathrm{t}},\mathrm{if}\mathrm{x}\to 1,\Rightarrow \mathrm{t}\to 0$$$$\mathrm{then}\mathrm{L}=-\mathrm{e}\underset{\mathrm{t}\to 0}{\lim }\frac{\mathrm{t}-{\mathrm{e}}^{\mathrm{t}}+1}{{({\mathrm{e}}^{\mathrm{t}}-1)}^2}=\mathrm{e}\underset{\mathrm{t}\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{t}}-\mathrm{t}-1}{{\mathrm{t}}^2}\times {\left(\frac{\mathrm{t}}{{\mathrm{e}}^{\mathrm{t}}-1}\right)}^2$$$$\mathrm{L}=\mathrm{e}\underset{\mathrm{t}\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{t}}-\mathrm{t}-1}{{\mathrm{t}}^2}=\frac{\mathrm{e}}{4}\underset{\mathrm{u}\to 0}{\lim }\frac{{\mathrm{e}}^{2\mathrm{u}}-2\mathrm{u}-1}{{\mathrm{u}}^2},[\mathrm{t}=2\mathrm{u}]$$$$\mathrm{L}-\frac{\mathrm{e}}{4}=\frac{\mathrm{e}}{4}\underset{\mathrm{u}\to 0}{\lim }\frac{{\mathrm{e}}^{2\mathrm{u}}-{\mathrm{u}}^2-2\mathrm{u}-1}{{\mathrm{u}}^2}=\frac{\mathrm{e}}{4}\underset{\mathrm{u}\to 0}{\lim }\frac{{\mathrm{e}}^{2\mathrm{u}}-{(\mathrm{u}+1)}^2}{{\mathrm{u}}^2}$$$$\mathrm{L}-\frac{\mathrm{e}}{4}=\frac{\mathrm{e}}{4}\underset{\mathrm{u}\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{u}}-\mathrm{u}-1}{{\mathrm{u}}^2}({\mathrm{e}}^{\mathrm{u}}+\mathrm{u}+1)$$$$\mathrm{L}-\frac{\mathrm{e}}{4}=\frac{1}{4}\mathrm{L}\times 2=\frac{1}{2}\mathrm{L}$$$$\mathrm{L}=\frac{\mathrm{e}}{2}$$

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