Question
Which of the following is correct for \[f\left( x \right) = \left\{ \begin{array}{l}
\left( {x - e} \right){2^{ - {2^{\left( {\frac{1}{{\left( {e - x} \right)}}} \right)}}}},\,\,\,x \ne e{\rm{ \,at\, }}x = e\\
\,\,\,\,\,\,\,\,\,\,\,\,\,0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = e\,\,\,\,
\end{array} \right.\]
A.
$$f\left( x \right)$$ is discontinuous at $$x = e$$
B.
$$f\left( x \right)$$ is differentiable at $$x = e$$
C.
$$f\left( x \right)$$ is non-differentiable at $$x = e$$
D.
none of these
Answer :
$$f\left( x \right)$$ is non-differentiable at $$x = e$$
Solution :
$$\eqalign{
& f\left( {{e^ + }} \right) = \mathop {\lim }\limits_{h \to 0} \left( {e + h - e} \right){2^{ - {2^{\frac{1}{{e - \left( {e - h} \right)}}}}}} \cr
& = \mathop {\lim }\limits_{h \to 0} \left( h \right){2^{ - {2^{ - \frac{1}{h}}}}} \cr
& = 0 \times 1 \cr
& = 0 \cr
& \left( {{\text{As for }}h \to 0,\, - \frac{1}{h} \to - \infty \Rightarrow {2^{ - \frac{1}{h}}} \to 0} \right) \cr
& f\left( {{e^ - }} \right) = \mathop {\lim }\limits_{h \to 0} \left( { - h} \right){2^{ - {2^{\frac{1}{h}}}}} \cr
& = 0 \times 0 \cr
& = 0 \cr
& {\text{Hence, }}f\left( x \right){\text{ is continuous at }}x = e \cr
& f'\left( {{e^ + }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {e + h} \right) - f\left( e \right)}}{h} \cr
& = \mathop {\lim }\limits_{h \to 0} \frac{{h \times {2^{ - {2^{ - \frac{1}{h}}}}} - 0}}{h} \cr
& = \mathop {\lim }\limits_{h \to 0} {2^{ - {2^{\frac{1}{h}}}}} \cr
& = 1 \cr
& f'\left( {{e^ - }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {e - h} \right) - f\left( 0 \right)}}{{ - h}} \cr
& = \mathop {\lim }\limits_{h \to 0} \frac{{\left( { - h} \right) \times {2^{ - {2^{ - \frac{1}{h}}}}} - 0}}{{ - h}} \cr
& = \mathop {\lim }\limits_{h \to 0} {2^{ - {2^{\frac{1}{h}}}}} \cr
& = 0 \cr
& {\text{Hence, }}f\left( x \right){\text{ is non - differentiable at }}x = e \cr} $$