Given:

$f(x)=\left\{\begin{array}{c}x, 0 \leq x<\frac{1}{2} \\ \frac{1}{2}, x=\frac{1}{2} \\ 1-x, \frac{1}{2}<x \leq 1\end{array}\right.$

We observe

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

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

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

$\therefore \lim _{x \rightarrow \frac{1}{2}^{-}} f(x)=\lim _{x \rightarrow \frac{1}{2}^{+}} f(x)=f\left(\frac{1}{2}\right)$

Hence, $f(x)$ is continuous at $x=\frac{1}{2}$.