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Question Number 144772 by phally last updated on 29/Jun/21

Answered by liberty last updated on 29/Jun/21

lim_(x→1) (((d/dx)(x^x −1))/((d/dx)(ln (1−(√(x^2 −1))))) consider (d/dx)(x^x −1)=x^x (ln x+1) and (d/dx)(ln (1−(√(x^2 −1))))=((−(x/( (√(x^2 −1)))))/(1−(√(x^2 −1)))) so lim_(x→1) ((x^x (ln x−1)(1−(√(x^2 −1)))((√(x^2 −1))))/(−x)) = 0

$$\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{x}} −\mathrm{1}\right)}{\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{ln}\:\left(\mathrm{1}−\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\right)\right.}\: \\ $$$$\mathrm{consider}\:\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{x}} −\mathrm{1}\right)=\mathrm{x}^{\mathrm{x}} \left(\mathrm{ln}\:\mathrm{x}+\mathrm{1}\right) \\ $$$$\mathrm{and}\:\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{ln}\:\left(\mathrm{1}−\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\right)\right)=\frac{−\frac{\mathrm{x}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}}}{\mathrm{1}−\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}}\: \\ $$$$\mathrm{so}\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{x}} \left(\mathrm{ln}\:\mathrm{x}−\mathrm{1}\right)\left(\mathrm{1}−\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\right)\left(\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\right)}{−\mathrm{x}} \\ $$$$=\:\mathrm{0}\: \\ $$

Commented by phally last updated on 29/Jun/21

thank brother

$$\:\:\:\:\mathrm{thank}\:\mathrm{brother} \\ $$

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