Q. The value of $\lim_{x \rightarrow 2}\frac{e^{3x - 6} - 1}{\sin{(2 - x)}}$ is
A) $\frac{3}{2}$
B) 3
C) -3
D) -1
Explanation
(c)
$\lim_{x \rightarrow 2}\frac{e^{3x - 6} - 1}{\sin{(2 - x)}} = \lim_{x \rightarrow 2}\frac{e^{3x - 6}(3)}{{- cos}{(2 - x)}}$
[using L’ Hospital’s rule]
$= - \frac{3e^{0}}{\cos 0} = - 3$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.