Question
Discuss the continuity of the following functions at the indicated point:
$\text{f}\text{(x)}=\begin{cases}\frac{\text{|x}^2-1|}{\text{x}-1},\text{for} & \text{x} \neq1\\2, &\text{for} \text{ x} = 1\end{cases} \text{at x}=1$

Answer

Given,
$\text{f}\text{(x)}=\begin{cases}\frac{\text{|x}^2-1|}{\text{x}-1}, & \text{x} \neq1\\2, &\text{ x} = 1\end{cases}$
$\Rightarrow\text{f}\text{(x)}=\begin{cases}\text{x}+1, & \text{x}< -1\\-\text{x}-1, & -1\leq \text{x} <0\\\text{x}+1,& \text{x}>1\\2,&\text{x}=1\end{cases}$
We obseve
$(\text{LHL at x}=1)=\lim\limits_{\text{x} \rightarrow 1^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(1-h)}$
$=\lim\limits_{\text{h} \rightarrow 0}-\text{(1-h)}-1=\lim\limits_{\text{h} \rightarrow 0}-2+\text{h}=-2$
And, $\text{f}(1)=2$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 1^-}\text{f}\text{(x)}\neq\text{f}(1)$
Hence, f(x) is discontinuous at x = 1

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