Question Number 163666 by Zaynal last updated on 09/Jan/22
$$\boldsymbol{\mathrm{lim}}\:_{\boldsymbol{{x}}\rightarrow\boldsymbol{\pi}} \:\:\left(\frac{\boldsymbol{{x}}^{\boldsymbol{\pi}} \:β\:\boldsymbol{\pi}^{\boldsymbol{{x}}} }{\boldsymbol{{x}}β\boldsymbol{\pi}}\right)\:=?? \\ $$$$\ll\mathrm{zaynal}\gg \\ $$
Answered by mahdipoor last updated on 09/Jan/22
$${hop}\Rightarrow\underset{{x}\rightarrow\pi} {\mathrm{lim}}\frac{\pi{x}^{\piβ\mathrm{1}} β\pi^{{x}} {ln}\pi}{\mathrm{1}}=\pi^{\pi} \left(\mathrm{1}β{ln}\pi\right) \\ $$
Answered by Ar Brandon last updated on 09/Jan/22
$$=\pi^{\pi} β\pi^{\pi} \mathrm{ln}\pi=\pi^{\pi} \mathrm{ln}\left(\frac{{e}}{\pi}\right) \\ $$
Commented by Zaynal last updated on 09/Jan/22
$$\mathrm{ok}\: \\ $$
Answered by alephzero last updated on 09/Jan/22
$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\frac{{x}^{\pi} β\pi^{{x}} }{{x}β\pi}\:=\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\frac{\left({x}^{\pi} β\pi^{{x}} \right)'}{\left({x}β\pi\right)'}\:= \\ $$$$=\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\frac{\left({x}^{\pi} \right)'β\left(\pi^{{x}} \right)'}{\left({x}\right)'β\left(\pi\right)'}\:=\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\frac{\pi{x}^{\piβ\mathrm{1}} β\mathrm{ln}\left(\pi\right)\pi^{{x}} }{\mathrm{1}β\mathrm{0}}\:= \\ $$$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\left(\pi{x}^{\piβ\mathrm{1}} β\mathrm{ln}\left(\pi\right)\pi^{{x}} \right)\:= \\ $$$$=\:\pi\pi^{\piβ\mathrm{1}} β\mathrm{ln}\left(\pi\right)\pi^{\pi} \:=\:\pi^{\pi} β\mathrm{ln}\left(\pi\right)\pi^{\pi} \:= \\ $$$$=\:\pi^{\pi} \left(\mathrm{1}β\mathrm{ln}\left(\pi\right)\right) \\ $$$$\Rightarrow\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\frac{{x}^{\pi} β\pi^{{x}} }{{x}β\pi}\:=\:\pi^{\pi} \left(\mathrm{1}β\mathrm{ln}\left(\pi\right)\right) \\ $$
Answered by manxsolar last updated on 16/Jan/22
$${regla}\:{L}'{Hospital} \\ $$$$\frac{\mathrm{0}}{\mathrm{0}}\rightarrow\:\:\:\:\:\:\:\:\:\:\:\:\:{lim}\:\:\:_{Γ\rightarrow\pi} \:\:\:\:\:\:\frac{{f}'}{{g}'} \\ $$$$ \\ $$