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