Continuity and Differentiability — Class 12 Maths Solution

We have $f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{\frac{{2{x^2} - 3x - 2}}{{x - 2}},}&{{\rm{ if }}x \ne 2}\\{5,}&{{\rm{ if }}x = 2}\end{array}} \right.$at $x = 2$

At $x = 2,$ ${\rm{LHL}} = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{2{x^2} - 3x - 2}}{{x - 2}}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{2{{(2 - h)}^2} - 3(2 - h) - 2}}{{(2 - h) - 2}}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{8 + 2{h^2} - 8h - 6 + 3h - 2}}{{ - h}}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{2{h^2} - 5h}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \frac{{h(2h - 5)}}{{ - h}} = 5$
${\rm{RHL}} = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{2{x^2} - 3x - 2}}{{x - 2}}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{2{{(2 + h)}^2} - 3(2 + h) - 2}}{{(2 + h) - 2}}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{8 + 2{h^2} + 8h - 6 - 3h - 2}}{h}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{2{h^2} + 5h}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{h(2h + 5)}}{h} = 5$
and $f(2) = 5$

therefore, ${\rm{LHL}} = {\rm{RHL}} = f(2)$

Therefore we can say that $f(x)$ is continuous at $x = 2$.