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  4. If f(x)=e- (1/x2)∀ x≠ 0 and f(0)=0, then f' (0) is

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

NTA AbhyasNTA Abhyas 2022

Solution:

$R f^{\prime}(0)=\displaystyle\lim _{h \rightarrow 0} \frac{e^{-\frac{1}{(0+h)^{2}}}-0}{h}=\displaystyle\lim _{h \rightarrow 0} \frac{1}{h} e^{-\frac{1}{h^{2}}}$
$=\displaystyle\lim _{h \rightarrow 0} \frac{\frac{1}{h}}{e^{\frac{1}{h^{2}}}}$
$=\displaystyle\lim _{h \rightarrow 0} \frac{-\frac{1}{h^{2}}}{e^{\frac{1}{h^{2}}} \cdot\left(-\frac{2}{h^{3}}\right)}$
$=\displaystyle\lim _{h \rightarrow 0} \frac{h}{2 e^{\frac{1}{h^{2}}}}=\frac{0}{\in f t y}=0$
$L f^{\prime}(0)=\displaystyle\lim _{h \rightarrow 0} \frac{e^{-\frac{1}{(0-h)^{2}}}-0}{-h}=\displaystyle\lim _{h \rightarrow 0}\left(-\frac{1}{h}\right) e^{-\frac{1}{h^{2}}}$
$=\displaystyle\lim _{h \rightarrow 0} \frac{-\frac{1}{h}}{e^{\frac{1}{h^{2}}}}$
$=\displaystyle\lim _{h \rightarrow 0} \frac{\frac{1}{h^{2}}}{e^{\frac{1}{h^{2}}}\left(-\frac{2}{h^{2}}\right)}$
$=\displaystyle\lim _{h \rightarrow 0} \frac{h}{-2 e^{\frac{1}{h^{2}}}}=\frac{0}{\in f t y}=0$
Hence, $f^{\prime}(0)=0$