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Question
Show that
\[f\left( x \right) = \begin{cases}1 + x^2 , if & 0 \leq x \leq 1 \\ 2 - x , if & x > 1\end{cases}\]
Solution
Given:
\[f\left( x \right) = \begin{cases}\begin{array}1 + x^2 , if 0 \leq x \leq 1 \\ 2 - x, if x > 1\end{array}\end{cases}\]
We observe
(LHL at x = 1) =
\[\lim_{x \to 1^-} f\left( x \right) = \lim_{h \to 0} f\left( 1 - h \right)\]
\[= \lim_{h \to 0} \left( 1 + \left( 1 - h \right)^2 \right) = \lim_{h \to 0} \left( 2 + h^2 - 2h \right) = 2\]
(RHL at x = 1) = \[\lim_{x \to 1^+} f\left( x \right) = \lim_{h \to 0} f\left( 1 + h \right)\]
\[= \lim_{h \to 0} \left( 2 - \left( 1 + h \right) \right) = \lim_{h \to 0} \left( 1 - h \right) = 1\]
\[\lim_{x \to 1^-} f\left( x \right) \neq \lim_{x \to 1^+} f\left( x \right)\]
Thus, f(x) is discontinuous at x = 1.
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