Solution: The function provided is

$f(x)=\left\{\begin{array}{ccc}x^{10}-1, & \text { if } & x \leq 1 \\ x^{2}, & \text { if } & x>1\end{array}\right.$

At $x=1$,
Left Hand Limit $=\lim _{x \rightarrow-1} f(x)=\lim _{x \rightarrow 1}\left(x^{10}-1\right)=0$

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

$f(1)=1^{10}-1=0$

Since, L.H.L. $\neq$ R.H.L.

As a result, $f(x)$ is discontinuous at $x=1$

So now, for $x=c<1 \lim _{x \rightarrow c}\left(x^{10}-1\right)=c^{10}-1=f(c)$ and for $x=c>1 \lim _{x \rightarrow c}\left(x^{2}\right)=c^{2}=f(1)$

Hence, $f(x)$ is a continuous for all $x \in R-\{1\}$

As a result, for all given function $x=1$ is a point of discontinuity.