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Question Number 47401 by Aditya789 last updated on 09/Nov/18

Commented by maxmathsup by imad last updated on 10/Nov/18

$$letA\left(x\right)={\left(cosx\right)}^{\frac{1}{sinx}}\Rightarrow A\left(x\right)=e^{\frac{1}{sinx}ln\left(cosx\right)}butcosx\sim 1-\frac{x^2}{2}(x\to 0)\Rightarrow$$$$ln\left(cosx\right)\sim ln(1-\frac{x^2}{2})\sim -\frac{x^2}{2}alsosinx\sim x\Rightarrow A\left(x\right)\sim e^{-\frac{x}{2}}\Rightarrow$$$${lim}_{x\to 0}A\left(x\right)=1.$$

Answered by ajfour last updated on 09/Nov/18

$$=\underset{x\to 0}{\lim }{[1-(1-\cos x)]}^{1/\sin x}$$$$=\underset{x\to 0}{\lim }{\left\{{[1-2\sin {}^2\frac{x}{2}]}^{-1/2\sin {}^2\left(x/2\right)}\right\}}^{-\frac{2\sin {}^2\left(x/2\right)}{2\sin \left(x/2\right)\cos \left(x/2\right)}}$$$$=\underset{x\to 0}{\lim }{e}^{-\tan \left(x/2\right)}=1.$$

Answered by rahul 19 last updated on 11/Nov/18

$$e^{\underset{x\to 0}{\lim }\left\{(\cos x-1)\times \frac{1}{\sin x}\right\}}=e^{\underset{x\to 0}{\lim }\{-\tan \frac{x}{2}\}}=e^0=1.$$

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