$ = \mathop {\lim }\limits_{h \to 0} \,\,(0 + h)\,\frac{{{e^{1/0 + h}} - {e^{ - 1/0 + h}}}}{{{e^{1/0 + h}} + {e^{ - 1/0 + h}}}}$
$ = \mathop {\lim }\limits_{h \to 0} \,\,h\,\,\frac{{{e^{1/h}} - {e^{ - 1/h}}}}{{{e^{1/h}} + {e^{ - 1/h}}}} =0$
and $f(0 - 0) = \mathop {\lim }\limits_{h \to 0} f(0 - h) = \mathop {\lim }\limits_{h \to 0} \,\, - h\,\,\frac{{{e^{ - 1/h}} - {e^{1/h}}}}{{{e^{ - 1/h}} + {e^{1/h}}}} = 0$
and $f(0) = 0$;
$\therefore \,\,\,f(0 + 0) = f(0 - 0) = f(0)$
Hence $f$ is continuous at $x = 0.$
At remaining points $f(x)$ is obviously continuous.
Thus it is everywhere continuous.
Again, $L\,f'(0) = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{f(0 - h) - f(0)}}{{ - h}}$
$ = \mathop {\lim }\limits_{h \to 0} \,\frac{{h\,.\,\frac{{{e^{ - 1/h}} - {e^{1/h}}}}{{{e^{ - 1/h}} + {e^{1/h}}}} - 0}}{{ - h}} = - 1$
$R\,f'(0) = \mathop {\lim }\limits_{h \to 0} \,\frac{{f\,(0 + h) - f\,(0)}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{h\,\,\frac{{{e^{1/h}} - {e^{ - 1/h}}}}{{{e^{1/h}} + {e^{ - 1/h}}}}}}{h} = 1$
$\because \,\,L\,\,f'(0) \ne R\,f'(0)$
$\therefore \,\,f$ is not differentiable at $x = 0$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.