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  • Evaluating <img alt="\mathrm{\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{\frac{1}{1-\cos x}}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx%20%5Crightar

Evaluating \mathrm{\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{\frac{1}{1-\cos x}}}

Option: 1

\mathrm{\frac{1}{e^{\frac{1}{3}}}}


Option: 2

0


Option: 3

\mathrm{e}


Option: 4

1


Answers (1)

best_answer

                                            \mathrm{\ln L=\lim _{x \rightarrow 0} \frac{\ln (\sin x / x)}{1-\cos x}}

Apply l'Hospital's rule: 

                     \mathrm{\ln L=\lim _{x \rightarrow 0} \frac{x(x \cos x-\sin x)}{x^2 \sin ^2 x}=\lim _{x \rightarrow 0} \frac{x^2(x \cos x-\sin x)}{x^3 \sin ^2 x}}

Now we have: 

                                   \mathrm{\begin{aligned} & \lim _{x \rightarrow 0} \frac{(x \cos x-\sin x)}{x^3} \\ & =\lim _{x \rightarrow 0} \frac{-\sin x}{3 x}=-\frac{1}{3} \end{aligned}}

Finally: 

                                    \mathrm{\ln L=1 \times-\frac{1}{3} \Longrightarrow L=\frac{1}{\sqrt[3]{e}}}

Posted by

Ajit Kumar Dubey

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