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Question Number 131769 by liberty last updated on 08/Feb/21

$$\underset{x\to \pi /2}{\lim }\frac{2^{-\cos \mathrm{x}}-1}{\mathrm{x}(\mathrm{x}-\frac{\pi }{2})}=?$$

Answered by bemath last updated on 08/Feb/21

$$\underset{x\to \pi /2}{\lim }\frac{1}{\mathrm{x}}.\underset{x\to \pi /2}{\lim }\frac{2^{\sin (\mathrm{x}-\frac{\pi }{2})}-1}{\mathrm{x}-\frac{\pi }{2}}=$$$$\frac{2}{\pi }.\underset{\mathrm{h}\to 0}{\lim }\frac{2^{\sin \mathrm{h}}-1}{\mathrm{h}}=\frac{2}{\pi }.\underset{\mathrm{h}\to 0}{\lim }\frac{\ln 2.2^{\sin \mathrm{h}}.\cos \mathrm{h}}{1}$$$$=\frac{2\ln 2}{\pi }={\pi }^{-1}.\ln 4$$

Answered by Dwaipayan Shikari last updated on 08/Feb/21

$$\underset{x\to \frac{\pi }{2}}{\lim }\frac{2^{-cos\left(x\right)}-1}{x(x-\frac{\pi }{2})}=\frac{2^{-sin(\frac{\pi }{2}-x)}-1}{x(x-\frac{\pi }{2})}=\frac{1-(\frac{\pi }{2}-x)log\left(2\right)-1}{\frac{\pi }{2}(x-\frac{\pi }{2})}$$$$=\frac{2log\left(2\right)}{\pi }=\frac{log\left(4\right)}{\pi }\underset{x\to 0}{\lim }{a}^x=1+xlog\left(a\right)$$

Answered by mathmax by abdo last updated on 08/Feb/21

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{2^{-\mathrm{cosx}}-1}{\mathrm{x}(\mathrm{x}-\frac{\pi }{2})}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{x}-\frac{\pi }{2}=\mathrm{t}(\mathrm{so}\mathrm{t}\to \mathrm{o})\mathrm{and}$$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}(\mathrm{t}+\frac{\pi }{2})=\frac{2^{\mathrm{sint}}-1}{\mathrm{t}(\mathrm{t}+\frac{\pi }{2})}\mathrm{we}\mathrm{have}{2}^{\mathrm{sint}}={\mathrm{e}}^{\mathrm{sint}\ln \left(2\right)}$$$$\mathrm{sint}\sim \mathrm{t}-\frac{{\mathrm{t}}^3}{6}\Rightarrow {\mathrm{e}}^{\ln \left(2\right)\mathrm{sint}}\sim {\mathrm{e}}^{\ln \left(2\right)(\mathrm{t}-\frac{{\mathrm{t}}^3}{6})}\sim 1+\ln \left(2\right)(\mathrm{t}-\frac{{\mathrm{t}}^3}{6})\Rightarrow$$$$\mathrm{f}(\mathrm{t}+\frac{\pi }{2})\sim \frac{\ln \left(2\right)(\mathrm{t}-\frac{{\mathrm{t}}^3}{6})}{\mathrm{t}(\mathrm{t}+\frac{\pi }{2})}=\frac{\ln \left(2\right)(1-\frac{{\mathrm{t}}^2}{6})}{\mathrm{t}+\frac{\pi }{2}}\to \frac{2\ln \left(2\right)}{\pi }(\mathrm{t}\to 0)\Rightarrow$$$${\lim }_{\mathrm{x}\to \frac{\pi }{2}}\mathrm{f}\left(\mathrm{x}\right)=\frac{2\ln \left(2\right)}{\pi }$$

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