Question

If $f\left(x\right)=e^{-\frac{1}{x^2}}\forall x\ne 0$ and $f\left(0\right)=0,$ then $f^{{}^{\prime }}\left(0\right)$ is

• A. not defined
• B. $1$
• C. $0$
• D. $2$

Answer 1

Answer: (C)

$R{f}^{\prime }(0)=\underset{h\to 0}{\lim }\frac{e^{-\frac{1}{(0+h{)}^2}}-0}{h}=\underset{h\to 0}{\lim }\frac{1}{h}{e}^{-\frac{1}{h^2}}$
$=\underset{h\to 0}{\lim }\frac{\frac{1}{h}}{e^{\frac{1}{h^2}}}$
$=\underset{h\to 0}{\lim }\frac{-\frac{1}{h^2}}{e^{\frac{1}{h^2}}\cdot \left(-\frac{2}{h^3}\right)}$
$=\underset{h\to 0}{\lim }\frac{h}{2e^{\frac{1}{h^2}}}=\frac{0}{\in fty}=0$
$L{f}^{\prime }(0)=\underset{h\to 0}{\lim }\frac{e^{-\frac{1}{(0-h{)}^2}}-0}{-h}=\underset{h\to 0}{\lim }\left(-\frac{1}{h}\right)e^{-\frac{1}{h^2}}$
$=\underset{h\to 0}{\lim }\frac{-\frac{1}{h}}{e^{\frac{1}{h^2}}}$
$=\underset{h\to 0}{\lim }\frac{\frac{1}{h^2}}{e^{\frac{1}{h^2}}\left(-\frac{2}{h^2}\right)}$
$=\underset{h\to 0}{\lim }\frac{h}{-2e^{\frac{1}{h^2}}}=\frac{0}{\in fty}=0$
Hence, $f^{\prime }(0)=0$