Question
If \[f\left( x \right) = \left\{ \begin{array}{l}
x{e^{ - \left( {\frac{1}{{\left| x \right|}} + \frac{1}{x}} \right)}},\,\,x \ne 0\\
0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0
\end{array} \right.\] Then $$f\left( x \right)$$ is-
A.
discontinuous every where
B.
continuous as well as differentiable for all $$x$$
C.
continuous for all $$x$$ but not differentiable at $$x =0$$
D.
neither differentiable nor continuous at $$x =0$$
Answer :
continuous for all $$x$$ but not differentiable at $$x =0$$
Solution :
$$\eqalign{
& f\left( 0 \right) = 0\,;\,\,f\left( x \right) = x{e^{ - \left( {\frac{1}{{\left| x \right|}} + \frac{1}{x}} \right)}} \cr
& {\text{R}}{\text{.H}}{\text{.L}}{\text{. }}\mathop {\lim }\limits_{h \to 0} \left( {0 + h} \right){e^{ - \frac{2}{h}}} = \mathop {\lim }\limits_{h \to 0} \frac{h}{{{e^{\frac{2}{h}}}}} = 0 \cr
& {\text{L}}{\text{.H}}{\text{.L}}{\text{. }}\mathop {\lim }\limits_{h \to 0} \left( {0 - h} \right){e^{ - \left( {\frac{1}{h}\, - \,\frac{1}{h}} \right)}} = 0 \cr} $$
Therefore, $$f\left( x \right)$$ is continuous.
$$\eqalign{
& {\text{R}}{\text{.H}}{\text{.D}}{\text{. }} = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {0 + h} \right){e^{ - \left( {\frac{1}{h}\, + \frac{1}{h}} \right)}} - 0}}{h} = 0 \cr
& {\text{L}}{\text{.H}}{\text{.D}}{\text{. }} = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {0 - h} \right){e^{ - \left( {\frac{1}{h}\, - \,\frac{1}{h}} \right)}} - 0}}{{ - h}} = 1 \cr} $$
Therefore, L.H.D. $$ \ne $$ R.H.D.
$$f\left( x \right)$$ is not differentiable at $$x =0.$$