If $f(x) = \left\{ \begin{array}{r} xe^{- \left( \frac{1}{|x|} + \frac{1}{x} ight)},\ \ x
eq 0 \ 0,\ \ \ x = 0 \end{array} ight.\$, then $f(x)$ is

Question

If $f(x) = \left\{ \begin{array}{r} xe^{- \left( \frac{1}{|x|} + \frac{1}{x} ight)},\ \ x 
eq 0 \ 0,\ \ \ x = 0 \end{array} ight.\$, then $f(x)$ is

Dr. Abubacker · Accepted Answer

(b)
LHL$= {\lim_{h \rightarrow 0}(0 - h)_{e}}^{- \left( \frac{1}{| - h|} + \frac{1}{( - h)} \right)} = \lim_{h \rightarrow 0}( - h) = 0$
RHL$= {\lim_{h \rightarrow 0}(0 + h)_{e}}^{- \left( \frac{1}{|h|} + \frac{1}{(h)} \right)} = \lim_{h \rightarrow 0}\frac{h}{e^{2/h}} = 0$
LHL=RHL$= f(0)$
Therefore, $f(x)$ is continuous for all $x$
Differentiability at $x = 0$
$Lf^{'}(0) = \lim_{h \rightarrow 0}\frac{( - h)e^{- \left( \frac{1}{h} - \frac{1}{h} \right)}}{( - h) - 0} = 1$
$Rf’(0) = \ \lim_{h \rightarrow 0}\frac{he^{- \left( \frac{1}{h} + \frac{1}{h} \right) - 0}}{h - 0}$
$= \lim_{h \rightarrow 0}\frac{1}{e^{2/h}} = 0$
$\Rightarrow \ \ Rf^{'}(0)Lf^{'}(0)$
Therefore, $f(x)$ is not differentiable at $x = 0$

Q. If $f(x) = \left\{ \begin{array}{r} xe^{- \left( \frac{1}{|x|} + \frac{1}{x} \right)},\ \ x \neq 0 \\ 0,\ \ \ x = 0 \end{array} \right.\$, then $f(x)$ is

Explanation

Reviewed by Dr. Abubacker

Want more practice?