Q. $\lim_{x \rightarrow 0}{x\log_{e}{(\sin x)}}$ is equal to

A) -1
B) $\log_{e}1$
C) 1
D) None of these
Explanation

(b)
$\lim_{x \rightarrow 0}{\log_{e}\left( \sin x \right)^{x}} = \log_{e}\left\lbrack \lim_{x \rightarrow 0}\left( \sin x \right)^{x} \right\rbrack$
$= \log_{e}\left\lbrack \lim_{x \rightarrow 0}\left( 1 + \sin x - 1 \right)^{\frac{x(\sin{x - 1})}{(\sin{x - 1})}} \right\rbrack$
$= \log_{e}\left\lbrack e^{\lim_{x \rightarrow 0}\ \ x(\sin{x - 1})} \right\rbrack$
$= \log_{e}1$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.