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Question Number 163666 by Zaynal last updated on 09/Jan/22
lim _(x→𝛑)   (((x^𝛑  βˆ’ 𝛑^x )/(xβˆ’π›‘))) =??  β‰ͺzaynal≫
$$\boldsymbol{\mathrm{lim}}\:_{\boldsymbol{{x}}\rightarrow\boldsymbol{\pi}} \:\:\left(\frac{\boldsymbol{{x}}^{\boldsymbol{\pi}} \:βˆ’\:\boldsymbol{\pi}^{\boldsymbol{{x}}} }{\boldsymbol{{x}}βˆ’\boldsymbol{\pi}}\right)\:=?? \\ $$$$\ll\mathrm{zaynal}\gg \\ $$
Answered by mahdipoor last updated on 09/Jan/22
hopβ‡’lim_(xβ†’Ο€) ((Ο€x^(Ο€βˆ’1) βˆ’Ο€^x lnΟ€)/1)=Ο€^Ο€ (1βˆ’lnΟ€)
$${hop}\Rightarrow\underset{{x}\rightarrow\pi} {\mathrm{lim}}\frac{\pi{x}^{\piβˆ’\mathrm{1}} βˆ’\pi^{{x}} {ln}\pi}{\mathrm{1}}=\pi^{\pi} \left(\mathrm{1}βˆ’{ln}\pi\right) \\ $$
Answered by Ar Brandon last updated on 09/Jan/22
=Ο€^Ο€ βˆ’Ο€^Ο€ lnΟ€=Ο€^Ο€ ln((e/Ο€))
$$=\pi^{\pi} βˆ’\pi^{\pi} \mathrm{ln}\pi=\pi^{\pi} \mathrm{ln}\left(\frac{{e}}{\pi}\right) \\ $$
Commented by Zaynal last updated on 09/Jan/22
ok
$$\mathrm{ok}\: \\ $$
Answered by alephzero last updated on 09/Jan/22
lim_(xβ†’Ο€) ((x^Ο€ βˆ’Ο€^x )/(xβˆ’Ο€)) = lim_(xβ†’Ο€) (((x^Ο€ βˆ’Ο€^x )β€²)/((xβˆ’Ο€)β€²)) =  = lim_(xβ†’Ο€) (((x^Ο€ )β€²βˆ’(Ο€^x )β€²)/((x)β€²βˆ’(Ο€)β€²)) = lim_(xβ†’Ο€) ((Ο€x^(Ο€βˆ’1) βˆ’ln(Ο€)Ο€^x )/(1βˆ’0)) =  lim_(xβ†’Ο€) (Ο€x^(Ο€βˆ’1) βˆ’ln(Ο€)Ο€^x ) =  = ππ^(Ο€βˆ’1) βˆ’ln(Ο€)Ο€^Ο€  = Ο€^Ο€ βˆ’ln(Ο€)Ο€^Ο€  =  = Ο€^Ο€ (1βˆ’ln(Ο€))  β‡’ lim_(xβ†’Ο€) ((x^Ο€ βˆ’Ο€^x )/(xβˆ’Ο€)) = Ο€^Ο€ (1βˆ’ln(Ο€))
$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\frac{{x}^{\pi} βˆ’\pi^{{x}} }{{x}βˆ’\pi}\:=\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\frac{\left({x}^{\pi} βˆ’\pi^{{x}} \right)'}{\left({x}βˆ’\pi\right)'}\:= \\ $$$$=\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\frac{\left({x}^{\pi} \right)'βˆ’\left(\pi^{{x}} \right)'}{\left({x}\right)'βˆ’\left(\pi\right)'}\:=\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\frac{\pi{x}^{\piβˆ’\mathrm{1}} βˆ’\mathrm{ln}\left(\pi\right)\pi^{{x}} }{\mathrm{1}βˆ’\mathrm{0}}\:= \\ $$$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\left(\pi{x}^{\piβˆ’\mathrm{1}} βˆ’\mathrm{ln}\left(\pi\right)\pi^{{x}} \right)\:= \\ $$$$=\:\pi\pi^{\piβˆ’\mathrm{1}} βˆ’\mathrm{ln}\left(\pi\right)\pi^{\pi} \:=\:\pi^{\pi} βˆ’\mathrm{ln}\left(\pi\right)\pi^{\pi} \:= \\ $$$$=\:\pi^{\pi} \left(\mathrm{1}βˆ’\mathrm{ln}\left(\pi\right)\right) \\ $$$$\Rightarrow\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\frac{{x}^{\pi} βˆ’\pi^{{x}} }{{x}βˆ’\pi}\:=\:\pi^{\pi} \left(\mathrm{1}βˆ’\mathrm{ln}\left(\pi\right)\right) \\ $$
Answered by manxsolar last updated on 16/Jan/22
regla Lβ€²Hospital  (0/0)β†’             lim   _(Γ—β†’Ο€)       ((fβ€²)/(gβ€²))
$${regla}\:{L}'{Hospital} \\ $$$$\frac{\mathrm{0}}{\mathrm{0}}\rightarrow\:\:\:\:\:\:\:\:\:\:\:\:\:{lim}\:\:\:_{Γ—\rightarrow\pi} \:\:\:\:\:\:\frac{{f}'}{{g}'} \\ $$$$ \\ $$

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