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Question Number 163961 by qaz last updated on 12/Jan/22

$$\underset{\mathrm{x}\to 1^{-}}{\lim }\frac{{\mathrm{e}}^{\frac{1}{{\mathrm{x}}^2-1}}}{\mathrm{x}-1}=?$$

Answered by bobhans last updated on 12/Jan/22

$$\mathrm{x}-1=\mathrm{u}\Rightarrow \mathrm{x}=\mathrm{u}+1$$$$=\underset{\mathrm{u}\to 0}{\lim }\frac{{\mathrm{e}}^{\frac{1}{{\mathrm{u}}^2+2\mathrm{u}}}}{\mathrm{u}}=\underset{\mathrm{u}\to 0}{\lim }\frac{{\mathrm{e}}^{\frac{1}{\mathrm{u}}}.{\mathrm{e}}^{\frac{1}{\mathrm{u}+2}}}{\mathrm{u}}$$$$=\sqrt{\mathrm{e}}\times \underset{\mathrm{u}\to 0}{\lim }\frac{{\mathrm{e}}^{\frac{1}{\mathrm{u}}}}{\mathrm{u}}=\sqrt{\mathrm{e}}\times \underset{\mathrm{y}\to \mathrm{\infty }}{\lim }{\mathrm{ye}}^{\mathrm{y}}$$$$=\mathrm{\infty }$$

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