5. continuity and differentiation — Maths STD 12 Science — Question

$\mathop {\lim }\limits_{x \to {3^ + }} \,f(x) = \mathop {\lim }\limits_{h \to 0} f(3 + h) = \mathop {\lim }\limits_{h \to 0} \,\,|3 + h - 3|\,\, = 0$

$\because \,\,\mathop {\lim }\limits_{x \to {3^ - }} \,f(x) = \mathop {\lim }\limits_{x \to {3^ + }} f(x) = f(3)$

Hence $f$ is continuous at $x = 3$

Now $L\,f'(3) = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{f(3 - h) - f(3)}}{{ - h}}$

$ = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{|3 - h - 3|\,\, - 0}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \,\frac{h}{{ - h}} = - 1$

$R\,f'(3) = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{f(3 + h) - f(3)}}{h}$

$ = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{|3 + h - 3|\,\, - 0}}{h} = 1$

$\because L\,{f}'(3)\,\ne \,R\,{f}'(3)$

Hence $f$ is not differentiable at $x = 3$.

Trick : Can be seen by graph it is continuous but tangent is not defined at $x = 3$.