Solution: The function provided is $f(x)= \begin{cases}x+5, & \text { if } \quad x \leq 1 \\ x-5, & \text { if } \quad x>1\end{cases}$
At $x=1$,
Left Hand Limit = $\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(x+5)=6$
Right Hand Limit $=\lim _{x \rightarrow-1} f(x)=\lim _{x \rightarrow 1}(x-5)=-4$
Since, L.H.L. $\neq$ R.H.L.
Hence, $f(x)$ is discontinuous at $x=1$
Now, for $x=c<1$
$\lim _{x \rightarrow 6}(x+5)=c+5=f(c)$ and
for $x=c>1 \lim _{x \rightarrow 6}(x-5)=c-5=f(c)$
Hence, $\mathrm{f}(\mathrm{x})$ is a continuous for all $\mathrm{x} \in \mathrm{R}-\{\mathbf{1}\}$
As a result, $\mathrm{f}(\mathrm{x})$ is not a continuous function.