Question

Prove that
$f\left(x\right)=\{\begin{array}{c}\frac{x^2-x-6}{x-3},whenx\ne 3\\ 5,whenx=3\end{array}$ is continuous at $x=3$.

Answer 1

L.H.L$={lim}_{x\to 3^{-}}\frac{x^2-x-6}{x-3}$

$={lim}_{x\to 3^{-}}\frac{x^2-3x+2x-6}{x-3}$

$={lim}_{x\to 3^{-}}\frac{x(x-3)+2(x-3)}{x-3}$

$={lim}_{x\to 3^{-}}\frac{(x-3)(x+2)}{x-3}$

$={lim}_{x\to 3^{-}}(x+2)$

$=3+2=5$

$f\left(3\right)=5$(given)

R.H.L$={lim}_{x\to 3^{+}}\frac{x^2-x-6}{x-3}$

$={lim}_{x\to 3^{+}}\frac{x^2-3x+2x-6}{x-3}$

$={lim}_{x\to 3^{+}}\frac{x(x-3)+2(x-3)}{x-3}$

$={lim}_{x\to 3^{+}}\frac{(x-3)(x+2)}{x-3}$

$={lim}_{x\to 3^{+}}(x+2)$

$=3+2=5$

$\therefore$ L.H.L$=$R.H.L$=5$ and $f\left(3\right)=5$

Hence the function is continuous at $x=3$