Question

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

• A. $2$
• B. $-1$
• C. $0$
• D. $1$

Answer 1

The correct option is D $1$

Given$f\left(x\right)=\frac{1}{x}-\frac{2}{e^{2x}-1}$

$\Rightarrow f\left(0\right)=\underset{x\to 0}{lim}\{\frac{1}{x}-\frac{2}{e^{2x}-1}\}=\underset{x\to 0}{lim}\frac{e^{2x}-1-2x}{x(e^{2x}-1)}.......\left[\frac{0}{0}form\right]$

$\therefore$ usingL'Hospital rule

$f\left(0\right)=\underset{x\to 0}{lim}\frac{2e^{2x}-2}{(e^{2x}-1+2x{e}^{2x})}$

$f\left(0\right)=\underset{x\to 0}{lim}\frac{4e^{2x}}{4{xe}^{2x}+2e^{2x}+2e^{2x}}=\frac{4.e^0}{4(0+e^0)}=1$