Q. Let $f(x) = \frac{\sqrt{x + 3}}{x + 1}$, then the value of $\lim_{x \rightarrow - 3 - 0}{f(x)}$ is
A) 0
B) Does not exist
C) $\frac{1}{2}$
D) $- \frac{1}{2}$
Explanation
(b)
$\lim_{x \rightarrow - 3 - 0} = \frac{\sqrt{x + 3}}{x + 1}$
But $\sqrt{x + 3}$ is not defined on left hand limit of -3.
Hence, function is not defined.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.