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Question Number 163666 by Zaynal last updated on 09/Jan/22

$$\mathbf{lim}{}_{\mathit{x}\to \mathit{\pi }}\left(\frac{{\mathit{x}}^{\mathit{\pi }}-{\mathit{\pi }}^{\mathit{x}}}{\mathit{x}-\mathit{\pi }}\right)=??$$$$\ll \mathrm{zaynal}\gg$$

Answered by mahdipoor last updated on 09/Jan/22

$$hop\Rightarrow \underset{x\to \pi }{\lim }\frac{\pi x^{\pi -1}-{\pi }^{x}ln\pi }{1}={\pi }^{\pi }(1-ln\pi )$$

Answered by Ar Brandon last updated on 09/Jan/22

$$={\pi }^{\pi }-{\pi }^{\pi }\ln \pi ={\pi }^{\pi }\ln \left(\frac{e}{\pi }\right)$$

Commented by Zaynal last updated on 09/Jan/22

$$\mathrm{ok}$$

Answered by alephzero last updated on 09/Jan/22

$$\underset{x\to \pi }{\lim }\frac{x^{\pi }-{\pi }^x}{x-\pi }=\underset{x\to \pi }{\lim }\frac{{(x^{\pi }-{\pi }^x)}^{\prime }}{{(x-\pi )}^{\prime }}=$$$$=\underset{x\to \pi }{\lim }\frac{{\left(x^{\pi }\right)}^{\prime }-{\left({\pi }^x\right)}^{\prime }}{{\left(x\right)}^{\prime }-{\left(\pi \right)}^{\prime }}=\underset{x\to \pi }{\lim }\frac{\pi x^{\pi -1}-\ln \left(\pi \right){\pi }^x}{1-0}=$$$$\underset{x\to \pi }{\lim }(\pi x^{\pi -1}-\ln \left(\pi \right){\pi }^x)=$$$$=\pi {\pi }^{\pi -1}-\ln \left(\pi \right){\pi }^{\pi }={\pi }^{\pi }-\ln \left(\pi \right){\pi }^{\pi }=$$$$={\pi }^{\pi }(1-\ln \left(\pi \right))$$$$\Rightarrow \underset{x\to \pi }{\lim }\frac{x^{\pi }-{\pi }^x}{x-\pi }={\pi }^{\pi }(1-\ln \left(\pi \right))$$

Answered by manxsolar last updated on 16/Jan/22

$$regla{L}^{\prime }Hospital$$$$\frac{0}{0}\to lim{}_{\times \to \pi }\frac{f^{\prime }}{g^{\prime }}$$$$$$

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