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Question
Discuss the continuity of the function f(x) at the point x = 1/2, where \[f\left( x \right) = \begin{cases}x, 0 \leq x < \frac{1}{2} \\ \frac{1}{2}, x = \frac{1}{2} \\ 1 - x, \frac{1}{2} < x \leq 1\end{cases}\]
Solution
Given:
\[f\left( x \right) = \begin{cases}x, 0 \leq x < \frac{1}{2} \\ \frac{1}{2}, x = \frac{1}{2} \\ 1 - x, \frac{1}{2} < x \leq 1\end{cases}\]
We observe
(LHL at x =\[\frac{1}{2}\]
\[\lim_{x \to \frac{1}{2}^-} f\left( x \right) = \lim_{h \to 0} f\left( \frac{1}{2} - h \right)\]
\[\lim_{h \to 0} \left( \frac{1}{2} - h \right) = \frac{1}{2}\]
(RHL at x = \[\frac{1}{2}\]
\[\lim_{x \to \frac{1}{2}^+} f\left( x \right) = \lim_{h \to 0} f\left( \frac{1}{2} + h \right)\]
\[\lim_{h \to 0} \left( 1 - \left( \frac{1}{2} + h \right) \right) = \frac{1}{2}\]
Also,
\[f\left( \frac{1}{2} \right) = \frac{1}{2}\]
\[\therefore \lim_{x \to \frac{1}{2}^-} f\left( x \right) = \lim_{x \to \frac{1}{2}^+} f\left( x \right) = f\left( \frac{1}{2} \right)\]
Hence,
\[f\left( x \right)\] is continuous at
\[x = \frac{1}{2}\] .
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