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Question Number 115604 by bemath last updated on 27/Sep/20

$$\left(1\right)\underset{x\to \frac{\pi }{2}}{\lim }\frac{\cos 2x}{\tan x}=?$$$$\left(2\right)\underset{x\to \pi }{\lim }\frac{\sqrt{2+\cos x}-1}{{(\pi -x)}^2}=?$$

Commented by Dwaipayan Shikari last updated on 27/Sep/20

$$\underset{x\to \pi }{\lim }\frac{2+\mathrm{cosx}-1}{{(\pi -\mathrm{x})}^2}.\frac{1}{\sqrt{2+\mathrm{cosx}}+1}=\frac{1}{4}\frac{{2\cos }^2\frac{\mathrm{x}}{2}}{{(\frac{\pi }{2}-\frac{\mathrm{x}}{2})}^2}.\frac{1}{2}$$$$=\frac{1}{4}.\frac{{\sin }^2(\frac{\pi }{2}-\frac{\mathrm{x}}{2})}{{(\frac{\pi }{2}-\frac{\mathrm{x}}{2})}^2}=\frac{1}{4}$$

Answered by bobhans last updated on 27/Sep/20

$$\left(2\right)\underset{x\to \pi }{\lim }\frac{\frac{-\sin x}{2\sqrt{2+\cos x}}}{-2(\pi -x)}=\underset{x\to \pi }{\lim }\frac{1}{4\sqrt{2+\cos x}}\times \underset{x\to \pi }{\lim }\frac{\sin x}{\pi -x}$$$$=\frac{1}{4}\times \underset{x\to \pi }{\lim }\frac{\cos x}{-1}=-\frac{1}{4}\times (-1)=\frac{1}{4}$$

Answered by Olaf last updated on 27/Sep/20

$$\left(2\right)$$$$x=\pi -u$$$$\underset{u\to 0}{\lim }\frac{\sqrt{2-\cos u}-1}{u^2}$$$$\underset{u\to 0}{\lim }\frac{\sqrt{2-(1-\frac{u^2}{2})}-1}{u^2}$$$$\underset{u\to 0}{\lim }\frac{\sqrt{1+\frac{u^2}{2}}-1}{u^2}$$$$\underset{u\to 0}{\lim }\frac{1+\frac{u^2}{4}-1}{u^2}$$$$\underset{u\to 0}{\lim }\frac{\frac{u^2}{4}}{u^2}=\frac{1}{4}$$$$$$$$$$

Answered by mathmax by abdo last updated on 27/Sep/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{\cos \left(2\mathrm{x}\right)}{\mathrm{tanx}}\mathrm{changement}\mathrm{t}=\mathrm{x}-\frac{\pi }{2}\mathrm{give}$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\cos \left(2(\mathrm{t}+\frac{\pi }{2})\right)}{\tan (\mathrm{t}+\frac{\pi }{2})}=-\frac{\cos \left(2\mathrm{t}\right)}{-\frac{1}{\mathrm{tant}}}=\mathrm{tant}.\cos \left(2\mathrm{t}\right)$$$$(\mathrm{x}\to \frac{\pi }{2}\Rightarrow \mathrm{t}\to 0)\Rightarrow \tan \left(\mathrm{t}\right)\sim \mathrm{t}\mathrm{and}\cos \left(2\mathrm{t}\right)\sim 1-{2\mathrm{t}}^2\Rightarrow$$$$\mathrm{tant}.\cos \left(2\mathrm{t}\right)\sim \mathrm{t}(1-{2\mathrm{t}}^2)\to 0\Rightarrow {\lim }_{\mathrm{x}\to \frac{\pi }{2}}\mathrm{f}\left(\mathrm{x}\right)=0$$

Answered by mathmax by abdo last updated on 27/Sep/20

$$2)\mathrm{let}\mathrm{g}\left(\mathrm{x}\right)=\frac{\sqrt{2+\mathrm{cosx}}-1}{{(\pi -\mathrm{x})}^2}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\pi -\mathrm{x}=\mathrm{t}\Rightarrow$$$$\mathrm{g}\left(\mathrm{x}\right)=\mathrm{g}(\pi -\mathrm{t})=\frac{\sqrt{2+\cos (\pi -\mathrm{t})}-1}{{\mathrm{t}}^2}=\frac{\sqrt{2-\mathrm{cost}}-1}{{\mathrm{t}}^2}$$$$\mathrm{we}\mathrm{have}\mathrm{cost}\sim 1-\frac{{\mathrm{t}}^2}{2}\Rightarrow -\mathrm{cost}\sim -1+\frac{{\mathrm{t}}^2}{2}\Rightarrow 2-\mathrm{cost}\sim 1+\frac{{\mathrm{t}}^2}{2}$$$$\Rightarrow \sqrt{2-\mathrm{cost}}\sim \sqrt{1+\frac{{\mathrm{t}}^2}{2}}\sim 1+\frac{{\mathrm{t}}^2}{4}\Rightarrow \mathrm{g}(\pi -\mathrm{t})\sim \frac{{\mathrm{t}}^2}{{4\mathrm{t}}^2}=\frac{1}{4}\Rightarrow$$$${\lim }_{\mathrm{t}\to 0}\mathrm{g}(\pi -\mathrm{t})=\frac{1}{4}={\lim }_{\mathrm{x}\to \pi }\mathrm{g}\left(\mathrm{x}\right)$$

Answered by bobhans last updated on 27/Sep/20

$$\left(1\right)setx=z+\frac{\pi }{2}$$$$\underset{z\to 0}{\lim }\frac{\cos (2z+\pi )}{\tan (z+\frac{\pi }{2})}=\underset{z\to 0}{\lim }\frac{-\cos 2z}{-\cot z}$$$$=\underset{z\to 0}{\lim }\tan z.\cos 2z=0.$$

Answered by Olaf last updated on 27/Sep/20

$$\underset{x\to \frac{\pi }{2}}{\lim }\cos 2x=-1$$$$\underset{x\to \frac{\pi }{2}}{\lim }\tan x=\pm \mathrm{\infty }$$$$\Rightarrow \underset{x\to \frac{\pi }{2}}{\lim }\frac{\cos 2x}{\tan x}=0$$$$$$$$$$

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