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. $f\left(x\right)$ is differentiable at $x=0$
• B. $f\left(0\right)=1$
• 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), (D)

$f^{\prime }\left(0\right)=\frac{-1}{3}$

${lim}_{h\to 0^{+}}f\left(x\right)={lim}_{h\to 0^{+}}f(0+h)={lim}_{h\to 0}[\frac{1}{0+h}-\frac{2}{e^{2(0+h)}-1}]$
$={lim}_{h\to 0}[\frac{1}{h}-\frac{2}{e^{2h}-1}]$
$={lim}_{h\to 0}\frac{e^{2h}-1-2h}{h(e^{2h}-1)}=1$
Similarly, ${lim}_{x\to 0^{-}}f\left(x\right)={lim}_{h\to 0}\frac{e^{-2h}-1+2h}{-h(e^{-2h}-1)}=1$
Hence $f\left(0\right)=1$
$R{f}^{\prime }\left(0\right)={lim}_{h\to 0}\frac{f(0+h)-f\left(0\right)}{h}={lim}_{h\to 0}\frac{\{\frac{1}{0+h}-\frac{2}{e^{2(0+h)}-1}\}-1}{h}$
$={lim}_{h\to 0}\frac{\frac{1}{h}-\frac{2}{e^{2h}-1}-1}{h}$
$=={lim}_{h\to 0}\frac{e^{2h}-1-2h-h(e^{2h}-1)}{h^2(e^{2h}-1)}=-\frac{1}{3}$
Similarly, $L{f}^{\prime }\left(0\right)=-\frac{1}{3}$
$\Rightarrow f\left(x\right)$ is differentiable at $x=0$ and $f^{\prime }\left(0\right)=-\frac{1}{3}$