Question

Find if the given below function
is continuous or discontinuous
at the indicated point
$f\left(x\right)=\{\begin{array}{cc}\frac{2x^2-3x-2}{x-2},& \text{if}x\ne 2\\ 5,& \text{if}x=2\end{array}\text{at}x=2$

Answer 1

Find left hand limit at $x=2$

Given that
$f\left(x\right)=\{\begin{array}{cc}\frac{2x^2-3x-2}{x-2},& \text{if}x\ne 2\\ 5,& \text{if}x=2\end{array}\text{at}x=2\dots \left(1\right)$
At $x=2,L.H.L=\underset{x\to 2^{-}}{\text{lim}}f\left(x\right)$
$=\underset{h\to 0^{+}}{\text{lim}}f(2-h)$
$=\underset{h\to 0^{+}}{\text{lim}}\frac{2(2-h{)}^2-3(2-h)-2}{(2-h)-2}$
$=\underset{h\to 0^{+}}{\text{lim}}\frac{8+2h^2-8h-6+3h-2}{-h}$
{$\because (a-b{)}^2=a^2+b^2-2ab$}
$=\underset{h\to 0^{+}}{\text{lim}}\frac{2h^2-5h}{-h}=\underset{h\to 0^{+}}{\text{lim}}\frac{h(2h-5)}{-h}=5\dots \left(2\right)$

Find right hand limit at $x=2$

$R.H.L=\underset{x\to 2^{+}}{\text{lim}}f\left(x\right)=\underset{h\to 0^{+}}{\text{lim}}f(2+h)$
$=\underset{h\to 0^{+}}{\text{lim}}\frac{2(2+h{)}^2-3(2+h)-2}{(2+h)-2}$
{$\because (a-b{)}^2=a^2+b^2-2ab$}
$=\underset{h\to 0^{+}}{\text{lim}}\frac{2h^2+5h}{h}$
$=\underset{h\to 0^{+}}{\text{lim}}\frac{h(2h+5)}{h}=5\dots \left(3\right)$

Also, $f\left(2\right)=5\dots \left(4\right)$
$\therefore L.H.L=R.H.L=f\left(2\right)$
[Using (2), (3), (4)]

$f\left(x\right)$ is continuous at $x=2$