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  • Evaluate    <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_%

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

Option: 1

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


Option: 2

0


Option: 3

1


Option: 4

\mathrm{e^{\frac{3}{5}}}


Answers (1)

best_answer

Let

                      \mathrm{f(x)=\left(\frac{\sin (x)}{x}\right)^{1 /(1-\cos (x))}}

For  \mathrm{|x|<\pi, \sin (x) / x>0}  and

                        \mathrm{\log (f(x))=\frac{\log (\sin (x) / x)}{1-\cos (x)}}

Now use Maclaurin series :

                         \mathrm{\frac{\sin (x)}{x}=1-\frac{x^2}{6}+\ldots}

So

                         \mathrm{\log \left(\frac{\sin (x)}{x}\right)=-\frac{x^2}{6}+\ldots}

while

                        \mathrm{1-\cos (x)=\frac{x^2}{2}+\ldots}

so

                        \mathrm{\lim _{x \rightarrow 0} \log (f(x))=-\frac{1}{3}}

and thus 

                       \mathrm{\lim _{x \rightarrow 0} f(x)=\exp (-1 / 3)}

 

Posted by

Gunjita

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