Question

A function f(x) is defined as,
$f\left(x\right)=\left\{\begin{array}{ll}\frac{x^2-x-6}{x-3};\mathrm{if}& x\ne 3\\ 5;\mathrm{if}& x=3\end{array}\right.$
Show that f(x) is continuous that x = 3.

Answer 1

Given:
$f\left(x\right)=\left\{\begin{array}{l}\frac{x^2-x-6}{x-3},x\ne 3\\ 5,x=3\end{array}\right.$

We observe
(LHL at x = 3) = $\underset{x\mathit{\to }{3}^{\mathit{-}}}{\lim }f\left(x\right)=\underset{h\mathit{\to }\mathit{0}}{\lim }f\left(3-h\right)$
=​$\underset{h\to 0}{\lim }\frac{{\left(3-h\right)}^2-\left(3-h\right)-6}{\left(3-h\right)-3}=\underset{h\to 0}{\lim }\frac{9+h^2-6h-3+h-6}{-h}=\underset{h\to 0}{\lim }\frac{h^2-5h}{-h}=\underset{h\to 0}{\lim }\left(5-h\right)=5$

And, (RHL at x = 3)​ = $\underset{x\mathit{\to }{3}^{+}}{\lim }f\left(x\right)=\underset{h\mathit{\to }0}{\lim }f\left(3+h\right)$
=​$\underset{h\to 0}{\lim }\frac{{\left(3+h\right)}^2-\left(3+h\right)-6}{\left(3+h\right)-3}=\underset{h\to 0}{\lim }\frac{9+h^2+6h-3-h-6}{h}=\underset{h\to 0}{\lim }\frac{h^2+5h}{h}=\underset{h\to 0}{\lim }\left(5+h\right)=5$

Also, $f\left(3\right)=5$

$\therefore \underset{x\mathit{\to }{3}^{+}}{\lim }f\left(x\right)=\underset{x\mathit{\to }{3}^{-}}{\lim }f\left(x\right)=f\left(3\right)$

Hence, $f\left(x\right)$ is continuous at $x=3$.