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

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

Commented by mohammad17 last updated on 26/Jun/21

$$??????$$

Answered by Olaf_Thorendsen last updated on 26/Jun/21

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

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

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

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