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Question Number 144544 by mohammad17 last updated on 26/Jun/21

lim_(x→(π/2)) (cosx)^(cotx)

$${lim}_{{x}\rightarrow\frac{\pi}{\mathrm{2}}} \left({cosx}\right)^{{cotx}} \\ $$

Commented by mohammad17 last updated on 26/Jun/21

??????

$$?????? \\ $$

Answered by Olaf_Thorendsen last updated on 26/Jun/21

f(x) = (cosx)^(cotx)   g(x) = f((π/2)−x) = (sinx)^(tanx)   lng(x) = tanx.ln(sinx)  lng(x) ∼_0  xln(x) → 0  ⇒ g(x) →_0  1  lim_(x→(π/2))  f(x) = lim_(x→0)  g(x) = 1

$${f}\left({x}\right)\:=\:\left(\mathrm{cos}{x}\right)^{\mathrm{cot}{x}} \\ $$$${g}\left({x}\right)\:=\:{f}\left(\frac{\pi}{\mathrm{2}}−{x}\right)\:=\:\left(\mathrm{sin}{x}\right)^{\mathrm{tan}{x}} \\ $$$$\mathrm{ln}{g}\left({x}\right)\:=\:\mathrm{tan}{x}.\mathrm{ln}\left(\mathrm{sin}{x}\right) \\ $$$$\mathrm{ln}{g}\left({x}\right)\:\underset{\mathrm{0}} {\sim}\:{x}\mathrm{ln}\left({x}\right)\:\rightarrow\:\mathrm{0} \\ $$$$\Rightarrow\:{g}\left({x}\right)\:\underset{\mathrm{0}} {\rightarrow}\:\mathrm{1} \\ $$$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:{f}\left({x}\right)\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{g}\left({x}\right)\:=\:\mathrm{1} \\ $$

Answered by mathmax by abdo last updated on 26/Jun/21

f(x)=(cosx)^(cotanx)  ⇒f(x)=e^(ctanxlog(cosx))   =_((π/2)−x=t)     e^(cotan((π/2)−t)log(sinx))  =e^(tant log(sint))    we have tant log(sint)=sintlog(sint)×(1/(cost))∼t logt→0 ⇒  lim_(x→(π/2)) f(x)=e^0  =1

$$\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{cosx}\right)^{\mathrm{cotanx}} \:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\mathrm{e}^{\mathrm{ctanxlog}\left(\mathrm{cosx}\right)} \\ $$$$=_{\frac{\pi}{\mathrm{2}}−\mathrm{x}=\mathrm{t}} \:\:\:\:\mathrm{e}^{\mathrm{cotan}\left(\frac{\pi}{\mathrm{2}}−\mathrm{t}\right)\mathrm{log}\left(\mathrm{sinx}\right)} \:=\mathrm{e}^{\mathrm{tant}\:\mathrm{log}\left(\mathrm{sint}\right)} \: \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{tant}\:\mathrm{log}\left(\mathrm{sint}\right)=\mathrm{sintlog}\left(\mathrm{sint}\right)×\frac{\mathrm{1}}{\mathrm{cost}}\sim\mathrm{t}\:\mathrm{logt}\rightarrow\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}} \mathrm{f}\left(\mathrm{x}\right)=\mathrm{e}^{\mathrm{0}} \:=\mathrm{1} \\ $$

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