Question
Show that $\text{f}\text{(x)}=\begin{cases}1+\text{x}^2,&\text{if } 0\leq\text{x}\leq 1\\2-\text{x},&\text{if }\text{x} > 1\end{cases}$ is discontinuous at x = 1.
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Answer
Given,
$\text{f}\text{(x)}=\begin{cases}1+\text{x}^2,&\text{if } 0\leq\text{x}\leq 1\\2-\text{x},&\text{if }\text{x} > 1\end{cases}$
We observe
$\text{(LHL at x}=1)\lim\limits_{\text{x} \rightarrow 1^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(1-\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}(1+1-\text{h)}^2=\lim\limits_{\text{h} \rightarrow 0}(2+\text{h}^2-\text{2h)}=2$
$\text{(RHL at x}=1)\lim\limits_{\text{x} \rightarrow 1^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(1+\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}(2-(1+\text{h))}=\lim\limits_{\text{h} \rightarrow 0}(1-\text{h)}=1$
$\lim\limits_{\text{x} \rightarrow 1^-}\text{f}\text{(x)}\neq\lim\limits_{\text{x} \rightarrow 1^+}\text{f}\text{(x)}$
Thus, f(x) is discontinuous at x = 1.
$\text{f}\text{(x)}=\begin{cases}1+\text{x}^2,&\text{if } 0\leq\text{x}\leq 1\\2-\text{x},&\text{if }\text{x} > 1\end{cases}$
We observe
$\text{(LHL at x}=1)\lim\limits_{\text{x} \rightarrow 1^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(1-\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}(1+1-\text{h)}^2=\lim\limits_{\text{h} \rightarrow 0}(2+\text{h}^2-\text{2h)}=2$
$\text{(RHL at x}=1)\lim\limits_{\text{x} \rightarrow 1^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(1+\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}(2-(1+\text{h))}=\lim\limits_{\text{h} \rightarrow 0}(1-\text{h)}=1$
$\lim\limits_{\text{x} \rightarrow 1^-}\text{f}\text{(x)}\neq\lim\limits_{\text{x} \rightarrow 1^+}\text{f}\text{(x)}$
Thus, f(x) is discontinuous at x = 1.
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