Question

Discuss the continuity of the function f(x) at the point x = 1/2, where \[f\left( x \right) = \begin{cases}x, 0 \leq x < \frac{1}{2} \\ \frac{1}{2}, x = \frac{1}{2} \\ 1 - x, \frac{1}{2} < x \leq 1\end{cases}\] 

Answer 1

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 

\[x = \frac{1}{2}\] .