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Question Number 14336 by tawa tawa last updated on 30/May/17

Commented by ajfour last updated on 30/May/17

$$\cos x=1-(1-\cos x)$$$$=1-2\sin {}^2\left(\frac{x}{2}\right)=1-t$$$$\underset{x\to 0}{\lim }\left(\cos x\right)=\underset{x\to 0}{\lim }{\left[{(1-t)}^{-1/t}\right]}^{-\mathit{t}\cot 2\mathit{x}}$$$$=\underset{x\to 0}{\lim }\left(e^{-t\cot 2x}\right)=\underset{x\to 0}{\lim }{\mathit{e}}^{-\mathit{y}}$$$$\underset{x\to 0}{\lim }y=\underset{x\to 0}{\lim }[t\cot 2x=\left(2\sin {}^2\frac{x}{2}\right)\left(\frac{\cos 2x}{2\sin \frac{x}{3}\cos \frac{x}{2}}\right)$$$$so\underset{x\to 0}{\lim }y=0$$$$hence\underset{x\to 0}{\lim }{\left(\cos x\right)}^{\cot 2x}=e^{-\underset{x\to 0}{\lim }y}$$$$=1.(perhapsonlyrighthand$$$$llimit).$$$$$$$$$$

Commented by tawa tawa last updated on 30/May/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 30/May/17

$$A={\left(cosx\right)}^{cot2x}$$$$lnA=cot2x.lncosx$$$$lim\left(l\underset{x\to 0}{nA}\right)=l\underset{x\to 0}{im}\frac{lncosx}{tgx}=\underset{x\to 0}{\lim }\frac{\frac{-sinx}{cosx}}{\frac{1}{{cos}^{2}x}}=$$$$=\underset{x\to 0}{\lim }(-sinx.cosx)=\underset{x\to 0}{\lim }\left(\frac{-1}{2}sin2x\right)=0$$$$lnA=0\Rightarrow A=\underset{x\to 0}{\lim }{\left(cosx\right)}^{cot2x}=1$$$$note:\underset{x\to 0}{\lim }tg2x=\underset{x\to 0}{\lim }tgx=0$$$$$$

Commented by tawa tawa last updated on 30/May/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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