Question

Check the continuity of the following function at $2$.
$f\left(x\right)=\{\begin{array}{ccc}\frac{1}{2}(x^2-4)& if& 0<x<2\\ 0& if& x=2\\ 2-8x^{-3}& if& x>2\end{array}$.

Answer 1

$RHL=\underset{x\to 2^{+}}{lim}2-8x^{-3}=1$

$LHL=\underset{x\to 2^{-}}{lim}\frac{x^2-4}{2}=0$

$f\left(x\right)$ at x = 2, we have 0

Hence, it is not continuous.