Question

Find all points of discontinuity of $f$, where $f$ is defined by
$f\left(x\right)=$ $\{\begin{array}{c}{x}^{10}-1,\text{if}x\le 1\\ x^2,\text{if}x>1\end{array}$

Answer 1

The given function is $f\left(x\right)=\{\begin{array}{c}{x}^{10}-1,\text{if}x\le 1\\ x^2,\text{if}x>1\end{array}$

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 $f\left(c\right)=c^{10}-1$ and $\underset{x\to c}{lim}f\left(x\right)=\underset{x\to c}{lim}(x^{10}-1)=c^{10}-1$
$\therefore$ $\underset{x\to c}{lim}f\left(x\right)=f\left(c\right)$

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

Case II :
If $c=1$, then the left hand limit of $x$ at $x=1$ is,

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

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

$\underset{x\to 1}{lim}f\left(x\right)=\underset{x\to 1}{lim}\left(x^2\right)=1^2=1$

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 $f\left(c\right)=c^2$

$\underset{x\to c}{lim}f\left(x\right)=\underset{x\to c}{lim}\left(x^2\right)=c^2$

$\therefore \underset{x\to c}{lim}f\left(x\right)=f\left(c\right)$

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 $f$.