$f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{3x + 5,}&{{\rm{ if }}x \ge 2}\\{{x^2},}&{{\rm{ if }}x < 2}\end{array}} \right.$at $x = 2$.}

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

At $x = 2,$ ${\rm{LHL}} = \mathop {\lim }\limits_{x \to {2^ - }} {(x)^2}$
$= \mathop {\lim }\limits_{h \to 0} {(2 - h)^2} = \mathop {\lim }\limits_{h \to 0} \left( {4 + {h^2} - 4h} \right) = 4$

and

${\rm{RHL}} = \mathop {\lim }\limits_{x \to {2^ + }} (3x + 5)$
$= \mathop {\lim }\limits_{h \to 0} [3(2 + h) + 5] = 11$

Since, ${\rm{LHL}} \ne {\rm{RHL}}$ at $x = 2$

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