Q. $\lim_{x \rightarrow 0}\left\lbrack \frac{e^{x} - e^{\sin x}}{x - \sin x} \right\rbrack$ is equal to
A) -1
B) 0
C) 1
D) None of these
Explanation
(c)
$\lim_{x \rightarrow 0}\left\lbrack \frac{e^{\sin x} - e^{x}}{\sin x - x} \right\rbrack = \lim_{x \rightarrow 0}\left\lbrack \frac{e^{x}(e^{\sin{x - x}} - 1)}{\sin x - x} \right\rbrack$
$= \lim_{x \rightarrow 0}\ e^{x}\ \lim_{x \rightarrow 0}\left\lbrack \frac{e^{\sin{x - x}} - 1}{\sin{x - x}} \right\rbrack$
$= e^{0} \times 1 = 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.