Let $f(x)=\left\{\begin{array}{cc}0, & x=0 \\ 2-x, & \text { for } 0 \lt x \lt 1 \\ 2, & \text { for } x=1…
Let $f(x)=\left\{\begin{array}{cc}0, & x=0 \\ 2-x, & \text { for } 0 \lt x \lt 1 \\ 2, & \text { for } x=1 \\ \frac{1}{2}-x, & \text { for } 1 \lt x \lt 2 \\ \frac{-3}{2}, & \text { for } x \geq 2\end{array}\right.$ then which of the following is true
$f$ is right continuous at $x=0$
$f$ is left continuous at $x=1$
$f$ is right continuous at $x=1$
$f$ is continuous at $x=2$
Solution
For $x=0$
$\begin{aligned}
& \lim _{x \rightarrow 0^{-}} f(x)=\text { does not exist as } f \text { is not defined for } x \lt 0 \\
& \lim _{x \rightarrow 0^{+}} f(x)=2, f(0)=0
\end{aligned}$
$f$ is not continuous for $x=0$
For $x=1 ; \lim _{x \rightarrow 1^{-}} f(x)=1, \lim _{x \rightarrow 1^{+}} f(x)=\frac{-1}{2}, f(1)=2$
So, $f$ is not continuous for $x=1$
For $x=2$;
$\lim _{x \rightarrow 2^{-}} f(x)=\frac{-3}{2}, \lim _{x \rightarrow 2^{+}} f(x)=\frac{-3}{2}, f(2)=\frac{-3}{2}$ So, $f$ is continuous at $x=2$.