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Question Number 79181 by naka3546 last updated on 23/Jan/20

$$\underset{x\to 0^{+}}{\lim }{(x^2+1)}^{\ln x}=...$$

Commented by john santu last updated on 23/Jan/20

$$\underset{x\to 0^{+}}{\lim }{\left\{{(1+x^2)}^{\frac{1}{x^2}}\right\}}^{x^{2}ln\left(x\right)}=$$$$e^{\underset{x\to 0^{+}}{\lim }ln\left(x^{x^2}\right)}=e^0=1$$

Answered by MJS last updated on 23/Jan/20

$$\underset{x\to 0^{+}}{\lim }\left(\ln x\ln (x^2+1)\right)=\underset{x\to 0^{+}}{\lim }\frac{\ln (x^2+1)}{\frac{1}{\ln x}}=$$$$\left[{\mathrm{L}}^{\prime }\mathrm{Hospital}\right]$$$$=\underset{x\to 0^{+}}{\lim }\frac{\frac{2x}{x^2+1}}{-\frac{1}{x{\left(\ln x\right)}^2}}=-2\underset{x\to 0^{+}}{\lim }\frac{{\left(x\ln x\right)}^2}{x^2+1}=0$$$$\Rightarrow$$$$\underset{x\to 0^{+}}{\lim }{(x^2+1)}^{\ln x}=1$$

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