Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{x(e^{1/x}-1)}{e^{1/x}+1}& x\ne 0\\ 0& x=0\end{array}$then
$f^{{}^{\prime }}\left(0^{+}\right)$ (or) the right derivative of $f$ at $x=0$

• A. $1$
• B. $-1$
• C. $2$
• D. Does not exist

Answer 1

The correct option is A $1$

$f\left(x\right)=\frac{x(e^{1/x}-1)}{e^{1/x}+1}$ when $x\ne 0$
$f^{\prime }\left(0^{+}\right)=\underset{x\to 0^{+}}{lim}\frac{f\left(x\right)-f\left(0\right)}{x-0}$
$f^{\prime }\left(0^{+}\right)=\underset{x\to 0^{+}}{lim}\frac{\frac{x(e^{1/x}-1)}{e^{1/x}+1}-0}{x}=\underset{x\to 0^{+}}{lim}\frac{e^{1/x}-1}{e^{1/x}+1}$
as $x\to 0^{+}\Rightarrow e^{-1/x}\to 0$
$f^{\prime }\left(0^{+}\right)=\underset{x\to 0^{+}}{lim}\frac{1-e^{-1/x}}{1+e^{-1/x}}=1$