Question

Check the continuity of the following function at ${}^{\prime }{2}^{\prime }$ $f\left(x\right)=\{\begin{array}{c}\frac{x^2-4}{2}if0<x<2\\ 0ifx=2\\ 2-8x^{-3}ifx>2\end{array}$

Answer 1

$\begin{array}{c}f\left(x\right)=\left\{\begin{array}{c}\frac{x^2-4}{2},if0<x<2\\ 0ifx=2\\ 2-8x^{3}ifx>2\end{array}\right\}\\ \Rightarrow f\left(2\right)=0\\ {lim}_{x\to 2}f\left(2\right)={lim}_{x\to 2}f\left(2-h\right)={lim}_{x\to 2}\frac{{\left(2-h\right)}^2}{2}-4={lim}_{x\to 2}\frac{h-4+h^2-4}{2}{lim}_{x\to 2}\frac{k^2-4h}{2}=0\\ {lim}_{x\to 2}f\left(2^{+}\right)={lim}_{x\to 2}f\left(2+h\right)={lim}_{x\to 2}2-\frac{8}{{\left(2+h\right)}^3}={lim}_{x\to 2}\frac{2\left(8+12h+6h^2+h^3\right)}{{\left(2+h\right)}^3}\\ {lim}_{x\to 2}\frac{16+24h+2h^2+2h^3-8}{{\left(2+h\right)}^3}=2{lim}_{x\to 2}\frac{\left(4+12h+6h^2+h^3\right)}{{\left(2+h\right)}^3}\\ =\therefore f\left(2^{-}\right)\ne f\left(2^{}\right)\end{array}$

It is discontinuous