Find all points of discontinuity of f, where f is defined by

Solution:

The given function $f$ is $f(x)= \begin{cases}x^{10}-1, & \text { if } x \leq 1 \\ x^{2}, & \text { if } x>1\end{cases}$

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

If $c<1$, then $f(c)=c^{10}-1$ and $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(x^{10}-1\right)=c^{10}-1$

$\therefore \lim _{x \rightarrow c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x<1$

Case II:

If $c=1$, then the left hand limit of $f$ at $x=1$ is,

$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}\left(x^{10}-1\right)=1^{10}-1=1-1=0$

The right hand limit of f at x = 1 is,

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

It is observed that the left and right hand limit of f at x = 1 do not coincide.

Therefore, f is not continuous at x = 1

Case III:

If $c>1$, then $f(c)=c^{2}$

$\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(x^{2}\right)=c^{2}$

$\therefore \lim _{x \rightarrow c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x>1$

Thus, from the above observation, it can be concluded that x = 1 is the only point of discontinuity of f.