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The given function can be rewritten as:

$f(x)=\left\{\begin{aligned} \frac{x-x}{2}, & \text { when } x>0 \\ \frac{x+x}{2}, & \text { when } x<0 \\ 2, & \text { when } x=0 \end{aligned}\right.$

$\Rightarrow f(x)=\left\{\begin{array}{l}0, \text { when } x>0 \\ x, \text { when } x<0 \\ 2, \text { when } x=0\end{array}\right.$

We observe

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

(RHL at $x=0)=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(h)=\lim _{h \rightarrow 0} 0=0$

And, $f(0)=2$

$\therefore \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x) \neq f(0)$

Thus, $f(x)$ is discontinuous at $x=0$.