Question

The function $f$ : $R/\left\{0\right\}\to R$ 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 $1$

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

Using expansion of $e^x$

$f\left(x\right)=\frac{(1+2x+\frac{{\left(2x\right)}^2}{2!}+\frac{{\left(2x\right)}^3}{3!}+...)-1-2x}{x{(e}^{2x}-1)}$

$f\left(x\right)=\frac{\frac{{\left(2x\right)}^2}{2!}+\frac{{\left(2x\right)}^3}{3!}}{x{(e}^{2x}-1)}=\frac{2x(\frac{\left(2x\right)}{2!}+\frac{{\left(2x\right)}^2}{3!})}{x\times {(e}^{2x}-1)}$

$f\left(x\right)=\frac{2x}{{(e}^{2x}-1)}\times \frac{(\frac{\left(2x\right)}{2!}+\frac{{\left(2x\right)}^2}{3!})}{x}$
Now,
$\underset{x\to 0}{lim}\frac{2x}{{(e}^{2x}-1)}\times \frac{\frac{\left(2x\right)}{2!}+\frac{{\left(2x\right)}^2}{3!}+...}{x}=1\times 1=1$
Hence, option D.