Question

The function $f\left(x\right)=\frac{1}{x}-\frac{2}{e^{2x}-1},x\ne 0$, is continuous at $x=0$. Then

• A. $f\left(0\right)=1$
• B. $f\left(x\right)$ is differentiable at $x=0$
• C. $f\left(x\right)$ is not differentiable at $x=0$
• D. $f^{{}^{\prime }}\left(0\right)=\frac{1}{3}$

Answer 1

Answer: (A), (B)

The correct options are
A $f\left(0\right)=1$
B $f\left(x\right)$ is differentiable at $x=0$

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

By L hospitals rule twice we get

$f\left(0\right)=\underset{x\to 0}{lim}\frac{4e^{2x}}{4e^{2x}+x4e^{2x}}=1$

For f(x) to be differentiable at x=0 $\underset{x\to 0}{lim}{f}^1\left(x\right)$ should exist i.e$\underset{x\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$ should exist.