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Evaluate \mathrm{\lim _{x \rightarrow 0}(\ln (x+e))^{\cot x}}

 

Option: 1

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


Option: 2

0


Option: 3

1


Option: 4

\mathrm{e}


Answers (1)

best_answer

Rewrite it as
\mathrm{ (\ln (x+e))^{\cot (x)}=\exp \left[x \cot (x) \frac{\ln (\ln (x+e))}{x}\right] }
And use the known limit

\mathrm{ \lim _{x \rightarrow 0} \frac{\sin (x)}{x}=1 }

To get
\mathrm{ \lim _{x \rightarrow 0} x \cot (x)=\lim _{x \rightarrow 0} \cos (x)\left(\frac{\sin (x)}{x}\right)^{-1}=1 \cdot 1^{-1}=1 }
and then use the definition of the derivative to compute

\mathrm{ \lim _{x \rightarrow 0} \frac{\ln (\ln (x+e))}{x}=\left.\frac{d}{d x} \ln (\ln (x+e))\right|_{x=0} }

Evaluating the derivative with chain rule and

\mathrm{ \frac{d}{d x} \ln (x)=\frac{1}{x} }

which should give you a final result of

\mathrm{ \lim _{x \rightarrow 0}(\ln (x+e))^{\cot (x)}=e^{1 / e} }

Hence option 1 is correct.

Posted by

sudhir.kumar

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