A function f(x) is defined as
Question:

A function f(x) is defined as

$f(x)=\left\{\begin{array}{rrr}\frac{x^{2}-9}{x-3} ; & \text { if } & x \neq 3 \\ 6 & ; & \text { if } x=3\end{array}\right.$

Show that $f(x)$ is continuous at $x=3$

Solution:

Given:

$f(x)= \begin{cases}\frac{x^{2}-9}{x-3}, & \text { if } x \neq 3 \\ 6, & \text { if } x=3\end{cases}$

We observe

$(\mathrm{LHL}$ at $x=3)=\lim _{x \rightarrow 3^{-}} f(x)=\lim _{h \rightarrow 0} f(3-h)$

$=\lim _{h \rightarrow 0} \frac{(3-h)^{2}-9}{(3-h)-3}=\lim _{h \rightarrow 0} \frac{3^{2}+h^{2}-6 h-9}{3-h-3}=\lim _{h \rightarrow 0} \frac{h^{2}-6 h}{-h}==\lim _{h \rightarrow 0} \frac{h(h-6)}{-h}=\lim _{h \rightarrow 0}(6-h)=6$

$(\mathrm{RHL}$ at $x=3)=\lim _{x \rightarrow 3^{+}} f(x)=\lim _{h \rightarrow 0} f(3+h)$

$=\lim _{h \rightarrow 0} \frac{(3+h)^{2}-9}{3+h-3}=\lim _{h \rightarrow 0} \frac{3^{2}+h^{2}+6 h-9}{h}=\lim _{h \rightarrow 0} \frac{h^{2}+6 h}{h}=\lim _{h \rightarrow 0} \frac{h(6+h)}{h}=\lim _{h \rightarrow 0}(6+h)=6$

Given:

$f(3)=6$

$\therefore \lim _{\mathrm{x} \rightarrow 3^{-}} f(x)=\lim _{\mathrm{x} \rightarrow 3^{+}} f(x)=f(3)$

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