Question

A function f(x) is defined as
$f\left(x\right)=\left\{\begin{array}{ll}\frac{x^2-9}{x-3};\mathrm{if}& x\ne 3\\ 6;\mathrm{if}& x=3\end{array}\right.$
Show that f(x) is continuous at x = 3

Answer 1

Given:
$f\left(x\right)=\left\{\begin{array}{l}\frac{x^2-9}{x-3},\mathrm{if}x\ne 3\\ 6,\mathrm{if}x=3\end{array}\right.$

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

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

Given:
$f\left(3\right)=6$


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

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