Question

Let $f\left(x\right)=\{\begin{array}{c}\frac{x}{1+e^{1/x}}x\ne 0,\\ 0x=0\end{array}.$ Then

• A. $f^{\prime }\left(0^{-}\right)=0$
• B. $f^{\prime }\left(0^{-}\right)=1$
• C. $f^{\prime }\left(0^{+}\right)=0$
• D. $f^{\prime }\left(0^{+}\right)=1$

Answer 1

Answer: (B), (C)

The correct options are
B $f^{\prime }\left(0^{-}\right)=1$
C $f^{\prime }\left(0^{+}\right)=0$

We know that,
$f^{\prime }\left(0^{-}\right)=\underset{x\to 0^{-}}{lim}\frac{f\left(x\right)-f\left(0\right)}{x}\Rightarrow f^{\prime }\left(0^{-}\right)=\underset{h\to 0^{+}}{lim}\frac{f(-h)-f\left(0\right)}{-h}\Rightarrow f^{\prime }\left(0^{-}\right)=\underset{h\to 0^{+}}{lim}\frac{\frac{-h}{1+e^{-1/h}}}{-h}\Rightarrow f^{\prime }\left(0^{-}\right)=\underset{h\to 0^{+}}{lim}\frac{1}{1+e^{-1/h}}=1$
And
$f^{\prime }\left(0^{+}\right)=\underset{x\to 0^{+}}{lim}\frac{f\left(x\right)-f\left(0\right)}{x}\Rightarrow f^{\prime }\left(0^{+}\right)=\underset{h\to 0^{+}}{lim}\frac{f\left(h\right)-f\left(0\right)}{h}\Rightarrow f^{\prime }\left(0^{+}\right)=\underset{h\to 0^{+}}{lim}\frac{\frac{h}{1+e^{1/h}}}{h}\Rightarrow f^{\prime }\left(0^{+}\right)=\underset{h\to 0^{+}}{lim}\frac{1}{1+e^{1/h}}=0$