Question

Discuss the continuity of the function f(x) at the point x = 1/2, where
$f\left(x\right)=\left\{\begin{array}{c}x,\\ 1/2,\\ 1-x,\end{array}\begin{array}{c}0\le x<1/2\\ x=1/2\\ 1/2<x\le 1\end{array}\right.$

Answer 1

Given:
$f\left(x\right)=\left\{\begin{array}{c}x,0\le x<\frac{1}{2}\\ \frac{1}{2},x=\frac{1}{2}\\ 1-x,\frac{1}{2}<x\le 1\end{array}\right.$

We observe

(LHL at x = $\frac{1}{2}$) = $\underset{x\mathit{\to }{\frac{1}{2}}^{-}}{\lim }f\left(x\right)=\underset{h\mathit{\to }0}{\lim }f\left(\frac{1}{2}-h\right)$ = ​$\underset{h\mathit{\to }0}{\lim }\left(\frac{1}{2}-h\right)=\frac{1}{2}$

(RHL at x = $\frac{1}{2}$) = $\underset{x\mathit{\to }{\frac{1}{2}}^{\mathit{+}}}{\lim }f\left(x\right)=\underset{h\mathit{\to }0}{\lim }f\left(\frac{1}{2}+h\right)$ = $\underset{h\mathit{\to }0}{\lim }\left(1-\left(\frac{1}{2}+h\right)\right)=\frac{1}{2}$

Also, ​$f\left(\frac{1}{2}\right)=\frac{1}{2}$


$\therefore \underset{x\mathit{\to }{\frac{1}{2}}^{-}}{\lim }f\left(x\right)=\underset{x\mathit{\to }{\frac{1}{2}}^{+}}{\lim }f\left(x\right)=f\left(\frac{1}{2}\right)$

Hence, $f\left(x\right)$ is continuous at $x=\frac{1}{2}$.