Solve this

Given: $f(x)=\left\{\begin{array}{l}\frac{x^{2}}{2}, \quad \text { if } 0 \leq \mathrm{x} \leq 1 \\ 2 x^{2}-3 x+\frac{3}{2}, \text { if } 1<\mathrm{x} \leq 2\end{array}\right.$

We have

$(\mathrm{LHL}$ at $x=1)=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0} \frac{(1-h)^{2}}{2}=\frac{1}{2}$

(RHL at $x=1)=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}\left[2(1+h)^{2}-3(1+h)+\frac{3}{2}\right]=2-3+\frac{3}{2}=\frac{1}{2}$

Also, $f(1)=\frac{(1)^{2}}{2}=\frac{1}{2}$

$\therefore \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(1)$

Hence, the given function is continuous at $x=1$.