Q. The function $f:R/\left\{ 0 \right\} \rightarrow R$ given by $f(x) = \frac{1}{x} - \frac{2}{e^{2x} - 1}$ Can be made continuous at $x = 0$ by defining $f(0)$ as function
A) 2
B) -1
C) 0
D) 1
Explanation
(d)
$\lim_{x \rightarrow 0}\frac{e^{2x} - 1 - 2x}{x(e^{2x} - 1)}$
$= \lim_{x \rightarrow 0}\frac{2e^{2x} - 2}{\left( e^{2x} - 1 \right) + 2xe^{2x}}$ [using L ‘Hospital rule]
$= \lim_{x \rightarrow 0}\frac{4e^{2x}}{4e^{2x} + 4xe^{2x}} = 1$ [using L ‘Hospital’s rule]
Since, $f(x)$ is continuous at $x = 0$, then
$\lim_{x \rightarrow 0}\ \ f(x) = f(0)\ \ \ \ \ \Rightarrow \ \ \ \ 1 = f(0)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.