$$f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{\frac{{{x^2}}}{2},}&{{\rm{ if }}0 \le x \le 1}\\{2{x^2} - 3x + \frac{3}{2},}&{{\rm{ if }}1 < x \le 2}\end{array}} \right.$$at $x = 1$ }

We have $$f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{\frac{{{x^2}}}{2},}&{{\rm{ if }}0 \le x \le 1}\\{2{x^2} - 3x + \frac{3}{2},}&{{\rm{ if }}1 < x \le 2}\end{array}} \right.$$at $x = 1$

At $x = 1,$ ${\rm{HL}} = \mathop {\lim }\limits_{x \to {1^ - }} \frac{{{x^2}}}{2} = \mathop {\lim }\limits_{h \to 0} \frac{{{{(1 - h)}^2}}}{2}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{1 + {h^2} - 2h}}{2} = \frac{1}{2}$

RHL $= \mathop {\lim }\limits_{x \to {1^ + }} \left( {2{x^2} - 3x + \frac{3}{2}} \right)$
$= \mathop {\lim }\limits_{h \to 0} \left[ {2{{(1 + h)}^2} - 3(1 + h) + \frac{3}{2}} \right]$
$= \mathop {\lim }\limits_{h \to 0} \left( {2 + 2{h^2} + 4h - 3 - 3h + \frac{3}{2}} \right) = - 1 + \frac{3}{2} = \frac{1}{2}$
and $f(1) = \frac{{{1^2}}}{2} = \frac{1}{2}$

$\Rightarrow {\rm{LHL}} = {\rm{RHL}} = f(1)$

Therefore we can say that $f(x)$ is continuous at $x = 1$.