# A function f(x) is defined as,

Question:

A function f(x) is defined as,

$f(x)=\left\{\begin{array}{rr}\frac{x^{2}-x-6}{x-3} ; & \text { if } \quad x \neq 3 \\ 5 & ; \text { if } x=3\end{array}\right.$

Show that f(x) is continuous that x = 3.

Solution:

Given:

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

We observe

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

$=\lim _{h \rightarrow 0} \frac{(3-h)^{2}-(3-h)-6}{(3-h)-3}=\lim _{h \rightarrow 0} \frac{9+h^{2}-6 h-3+h-6}{-h}=\lim _{h \rightarrow 0} \frac{h^{2}-5 h}{-h}=\lim _{h \rightarrow 0}(5-h)=5$

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

$=\lim _{h \rightarrow 0} \frac{(3+h)^{2}-(3+h)-6}{(3+h)-3}=\lim _{h \rightarrow 0} \frac{9+h^{2}+6 h-3-h-6}{h}=\lim _{h \rightarrow 0} \frac{h^{2}+5 h}{h}=\lim _{h \rightarrow 0}(5+h)=5$

Also, $f(3)=5$

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

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