Q. For the function$f(x) = \frac{e^{1/x} - 1}{e^{1/x} + 1},\ x = 0$, which of the following is correct?
A) $\lim_{x \rightarrow 0}{f(x)}$ does not exist
B) $\lim_{x \rightarrow 0}{f(x)} = 1$
C) $\lim_{x \rightarrow 0}{f(x)}$ exists but $f(x)$ is not continuous at $x = 0$
D) $f(x)$ is continuous at $x = 0$
Explanation
(a)
LHL$= \lim_{h \rightarrow 0}{f(0 - h)}$
$= \lim_{h \rightarrow 0}\frac{e^{- 1/h} - 1}{e^{- 1/h} + 1} = - 1$ $\left\lbrack \because\ \lim_{h \rightarrow 0}\ \frac{1}{e^{1/h}} = 0\ \ \right\rbrack$
RHL$= \lim_{h \rightarrow 0}f(0 + h) = \lim_{h \rightarrow 0}\frac{e^{1/h} - 1}{e^{1/h} + 1}$
$= \lim_{h \rightarrow 0}\frac{1 - \frac{1}{e^{1/h}}}{1 + \frac{1}{e^{1/h}}} = 1$
$\therefore$ LHL$\neq$RHL
So, limit does not exist at $x = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.