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Question Number 128604 by john_santu last updated on 08/Jan/21

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

Answered by liberty last updated on 09/Jan/21

$$\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{\mathrm{x}}\right)={\mathrm{x}}^{\mathrm{x}}(\ln \mathrm{x}+1)$$$$\underset{x\to 1}{\lim }\frac{{\mathrm{x}}^{\mathrm{x}}(\ln \mathrm{x}+1)-1}{{\mathrm{x}}^{-1}-1}=\underset{x\to 1}{\lim }\frac{{\mathrm{x}}^{\mathrm{x}}.\ln \mathrm{x}+{\mathrm{x}}^{\mathrm{x}}-1}{{\mathrm{x}}^{-1}-1}$$$$=\underset{x\to 1}{\lim }\frac{{\mathrm{x}}^{\mathrm{x}}{(\ln \mathrm{x}+1)}^2+{\mathrm{x}}^{\mathrm{x}-1}}{-{\mathrm{x}}^{-2}}$$$$=\frac{1+1}{-1}=-2$$

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