Question

Let the function $f\left(x\right)=\frac{x}{1+e^{1/x}}$ if $x\ne 0,f\left(x\right)=0$. If $x=0$, the derivative from the right $f^{\prime }\left(0^{+}\right)$ and the derivative from the left $f^{\prime }\left(0^{-}\right)$ are

• A. $0$ and $1$
• B. $-1$ and $-1$
• C. $1$ and $-1$
• D. $1$ and $0$

Answer 1

Answer: (A)

$0$ and $1$

$f^{\prime }\left(0^{+}\right)=\underset{h\to 0^{+}}{lim}\frac{f(0+h)-f\left(0\right)}{h}$

$=\underset{h\to 0^{+}}{lim}\frac{h}{h(1+e^{1/h})}=\underset{h\to 0^{+}}{lim}\frac{1}{1+e^{1/h}}=0$

$f^{\prime }\left(0^{+}\right)=\underset{h\to 0^{-}}{lim}\frac{f\left(h\right)-f\left(0\right)}{h}$

$=\underset{h\to 0^{-}}{lim}\frac{h}{h(1+e^{-1/h})}=\underset{h\to 0^{-}}{lim}\frac{1}{1+e^{-1/h}}=1$