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Question Number 193117 by mustafazaheen last updated on 04/Jun/23

$$\underset{x\to 0}{\lim }{\left(\mathrm{cosx}\right)}^{\frac{1}{\mathrm{x}}}$$

Answered by Subhi last updated on 04/Jun/23

$$y={lim}_{x\to 0}{\left(cosx\right)}^{\frac{1}{x}}$$$$ln\left(y\right)={lim}_{x\to 0}\frac{ln\left(cosx\right)}{x}$$$$applyL{Hopital}^{\prime }slaw$$$$=\frac{-sin\left(x\right)}{cos\left(x\right)}=-tan\left(x\right)=-tan\left(0\right)=0$$$$ln\left(y\right)=0$$$$y=e^0=1$$

Answered by horsebrand11 last updated on 04/Jun/23

$$\mathrm{L}={\mathrm{e}}^{\underset{x\to 0}{\lim }\left(\frac{\cos \mathrm{x}-1}{\mathrm{x}}\right)}$$$$\mathrm{L}={\mathrm{e}}^{\underset{x\to 0}{\lim }\left(\frac{-\sin {}^2\mathrm{x}}{\mathrm{x}(\cos \mathrm{x}+1)}\right)}={\mathrm{e}}^0=1$$

Answered by aba last updated on 04/Jun/23

$$={\underset{\mathrm{x}\to 0}{\lim e}}^{\frac{\ln (1-\frac{{\mathrm{x}}^2}{2}+o\left({\mathrm{x}}^2\right))}{\mathrm{x}}}$$$$={\underset{\mathrm{x}\to 0}{\lim e}}^{\frac{-\frac{{\mathrm{x}}^2}{2}+o\left({\mathrm{x}}^2\right)}{\mathrm{x}}}$$$$={\underset{x\to 0}{\lim e}}^{-\frac{\mathrm{x}}{2}+o\left(\mathrm{x}\right)}$$$$=\underset{x\to 0}{\lim }(1-\frac{\mathrm{x}}{2}+o\left(\mathrm{x}\right))=1$$

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