Solution: The function provided is

$f(x)=\left\{\begin{array}{lll}x+1, & \text { if } & x \geq 1 \\ x^{2}+1, & \text { if } & x<1\end{array}\right.$

As we all know that, $\mathrm{f}(\mathrm{x})$ being a polynomial is continuous for $x \geq 1$ and $x<1$ for all $x \in \mathrm{R}$.

Now check Continuity at $x=1$

Right Hand Limit =
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{1}}(x+1)=\lim _{k \rightarrow 0}(1+h+1)=2$

Left Hand Limit =
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}\left(x^{2}+1\right)=\lim _{h \rightarrow 0}\left((1-h)^{2}+1\right)=2$

And $f(1)=2$

Since, L.H.L. $=$ R.H.L. $=f(1)$

Hence, $f(x)$ is a continuous at $x=1$ for all $x \in R$. As a result, $f(x)$ has no point of discontinuity.