Answer
Function $f\left( x \right)=\left\{ \begin{align}
& -1\text{ if }x<0 \\
& \text{ }1\text{ if }x\ge 0 \\
\end{align} \right.$is discontinuous at $x=0$.
Work Step by Step
To find the number at which the provided function is discontinuous, check the continuity of the function at $x=0$.
Consider the provided function ,
$f\left( x \right)=\left\{ \begin{align}
& -1\text{ if }x<0 \\
& \text{ }1\text{ if }x\ge 0 \\
\end{align} \right.$
Condition 1: Check the provided function $f\left( x \right)$ is defined at $x=a=0$.
Since, $f\left( x \right)=f\left( a \right)=1$ at $x=0$.
Hence, the function $f\left( x \right)$ is defined at $a=0$.
Condition 2: Check the limit $\underset{x\to a}{\mathop{\lim }}\,f(x)$ exists in the function $f\left( x \right)$.
Consider the left-hand limit,
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right),\text{ if }f\left( x \right)=-1$
Therefore, the left-hand limit is $-1$.
Consider the right-hand limit,
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right),\text{ if }f\left( x \right)=1$
Therefore, the left-hand limit is $1$.
Since the left hand limit and right-hand limit are not equal to each other, the limit does not exist at $x=a=0$.
There is no need to check condition (3).
Hence, the function is discontinuous at $x=0$.
