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Question Number 131962 by EDWIN88 last updated on 10/Feb/21

$$\underset{x\to 1^{+}}{\lim }(\frac{1}{\ln \mathrm{x}}-\frac{1}{\mathrm{x}-1})=?$$

Answered by liberty last updated on 10/Feb/21

$$\mathrm{let}\mathrm{u}=\ln \mathrm{x}\Rightarrow \mathrm{x}={\mathrm{e}}^{\mathrm{u}}$$$$\underset{\mathrm{u}\to 0}{\lim }(\frac{1}{\mathrm{u}}-\frac{1}{{\mathrm{e}}^{\mathrm{u}}-1})$$$$=\underset{\mathrm{u}\to 0}{\lim }\left(\frac{{\mathrm{e}}^{\mathrm{u}}-1-\mathrm{u}}{\mathrm{u}({\mathrm{e}}^{\mathrm{u}}-1)}\right)$$$$=\underset{\mathrm{u}\to 0}{\lim }\left(\frac{{\mathrm{e}}^{\mathrm{u}}-1}{{\mathrm{e}}^{\mathrm{u}}-1+{\mathrm{ue}}^{\mathrm{u}}}\right)$$$$=\underset{\mathrm{u}\to 0}{\lim }\left(\frac{{\mathrm{e}}^{\mathrm{u}}}{{\mathrm{e}}^{\mathrm{u}}+{\mathrm{e}}^{\mathrm{u}}+{\mathrm{ue}}^{\mathrm{u}}}\right)=\frac{1}{2}$$

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