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Question Number 126056 by TITA last updated on 16/Dec/20

Commented by TITA last updated on 16/Dec/20

$$\mathrm{please}\mathrm{help}$$

Answered by Olaf last updated on 16/Dec/20

$$\frac{x^x-1}{\ln (x+1)}=\frac{e^{x\ln x}-1}{\ln (x+1)}$$$$\mathrm{La}\mathrm{limite}\mathrm{en}{0}^{+}\mathrm{est}\mathrm{de}\mathrm{la}\mathrm{forme}\frac{0^{+}}{0^{+}}.$$$${\mathrm{C}}^{\prime }\mathrm{est}\mathrm{une}\mathrm{forme}\mathrm{indeterminee}.$$$$\mathrm{On}\mathrm{utilise}\mathrm{la}\mathrm{regle}\mathrm{de}{\mathrm{l}}^{\prime }\mathrm{hopital}:$$$$\underset{x\to 0^{+}}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to 0^{+}}{\lim }\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}$$$$\mathrm{donc}\mathrm{on}\mathrm{cherche}:$$$$\underset{x\to 0^{+}}{\lim }\frac{(1.\ln x+x.\frac{1}{x})e^{x\ln x}}{1/(x+1)}$$$$\underset{x\to 0^{+}}{\lim }(x+1)(\ln x+1)e^{x\ln x}$$$$\mathrm{de}\mathrm{la}\mathrm{forme}1\times (-\mathrm{\infty })\times 1\mathrm{donc}-\mathrm{\infty }$$

Answered by Dwaipayan Shikari last updated on 16/Dec/20

$$\underset{x\to 0^{+}}{\lim }\frac{e^{xlogx}-1}{log(x+1)}=\frac{1+xlog\left(x\right)-1}{x}\underset{x\to 0}{\lim }log(x+1)=xand{e}^x=x+1$$$$=log\left(x\right)\to -\mathrm{\infty }$$

Answered by 676597498 last updated on 17/Dec/20

$$asx\to 0$$$$ln(x+1)\to x-\frac{x^2}{2!}+\xi \left(x\right)$$$$\underset{x\to 0^{+}}{\lim }\frac{x^x-1}{ln(x+1)}=\underset{x\to 0^{+}}{\lim }\frac{x^x-1}{x(1-\frac{x}{2})}$$$$=\underset{x\to 0^{+}}{\lim }\frac{x^{x-1}-\frac{1}{x}}{1-\frac{x}{2}}=-\mathrm{\infty }$$

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