Question

If $f\left(x\right)=\{\begin{array}{cc}x\frac{e^{\left(\frac{1}{x}\right)}-e^{\left(\frac{-1}{x}\right)}}{e^{\left(\frac{1}{x}\right)}+e^{\left(\frac{-1}{x}\right)}},& x\ne 0\\ 0,& x=0\end{array}$ then which of the following is

• A. f is continuous and differentiable at every point
• B. f is continuous at every point but is not differentiable
• C. f is differentiable at every point
• D. f is differentiable only at the origin

Answer 1

The correct option is B f is continuous at every point but is not differentiable

$f(0+0)={lim}_{h\to 0}f\left(x\right)={lim}_{h\to 0}f(0+h)$
$={lim}_{h\to 0}(0+h)\frac{e^{\frac{1}{0+h}}-e^{\frac{-1}{0+h}}}{e^{\frac{1}{0+h}}+e^{\frac{-1}{0+h}}}={lim}_{h\to 0}h\frac{e^{\frac{1}{h}}-e^{\frac{-1}{h}}}{e^{\frac{1}{h}}+e^{\frac{-1}{h}}}=0$
$andf(0-0)={lim}_{h\to 0}f(0-h)={lim}_{h\to 0}-h\frac{e^{\frac{-1}{h}}-e^{\frac{1}{h}}}{e^{\frac{-1}{h}}+e^{\frac{1}{h}}}=0$
$andf\left(0\right)=0;\therefore f(0+0)=f(0-0)=f\left(0\right)$
Hence f is continuous at x = 0.
At remaining points f(x) is obviously continuous.
Thus it is everywhere continuous.
Again, $L{f}^{\prime }\left(0\right)={lim}_{h\to 0}\frac{f(0-h)-f\left(0\right)}{-h}$
$={lim}_{h\to 0}\frac{-h.\frac{e^{\frac{-1}{h}}-e^{\frac{1}{h}}}{e^{\frac{-1}{h}}+e^{\frac{1}{h}}}}{-h}=-1$
$R{f}^{\prime }\left(0\right)={lim}_{h\to 0}\frac{f(0+h)-f\left(0\right)}{h}={lim}_{h\to 0}\frac{h\frac{e^{\frac{1}{h}}-e^{\frac{-1}{h}}}{e^{\frac{1}{h}}+e^{\frac{-1}{h}}}}{h}=1$
$\because L{f}^{\prime }\left(0\right)\ne R{f}^{\prime }\left(0\right)$
$\therefore f$ is not differentiable at x = 0