Question

The function $f : R /\{0\} \to R$ given by $f(x) = \frac{1}{x} - \frac{2}{e^{2x} - 1}$ can be made continuous at x = 0 by defining f (0) as

• A. 0
• B. 1
• C. 2
• D. -1

Answer 1

Answer: (B)

Given, $f\left(x\right) = \frac{1}{x} - \frac{2}{e^{2x}-1}$
$ \Rightarrow f\left(0\right) = \displaystyle\lim_{x \to0} \frac{1}{x} - \frac{2}{e^{2x} -1}$
$ = \displaystyle\lim_{x \to0} \frac{\left(e^{2x } -1\right)-2x}{x\left(e^{2x} -1\right)} $ [$\frac{0}{0}$ from]
$\therefore $ using, L'Hospital rule
$f\left(0\right) = \displaystyle\lim_{x \to0} \frac{4e^{2x}}{2\left(xe^{2x} 2+e^{2x }.1\right) +e^{2x}.2} $
$= \displaystyle\lim _{x \to 0} \frac{4e^{2x}}{4xe^{2x} + 2e^{2x} + 2e^{2x}}$ [$\frac{0}{0}$ from]
$= \displaystyle\lim_{x \to0} \frac{4e^{2x}}{4\left(xe^{2x} + e^{2x}\right)} = \frac{4.e^{0}}{4\left(0+e^{0}\right)} = 1 $