#### Need solution for RD Sharma Maths Class 12 Chapter 8 Continuity Excercise 8.1 Question 2

$x=3$

Hint:

For a function to be continuous at a point, its LHL RHL and value at that point should be equal.

Solution:

Given,

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

We observe,

[LHL at $x=3$ ]

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

$\lim _{h \rightarrow 0} \frac{(3-h)^{2}-(3-h)-6}{(3-h)-3}=\lim _{h \rightarrow 0} \frac{9+h^{2}-6 h-3+h-6}{-h}$                     $\left[\because(a-b)^{2}=\left(a^{2}+b^{2}-2 a b\right)\right]$

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{h^{2}+5 h}{h} \\\\ &=\lim _{h \rightarrow 0}(5+h)=5 \end{aligned}

[RHL at $x=3$ ]

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

$\lim _{h \rightarrow 0} \frac{(3+h)^{2}-(3+h)-6}{(3+h)-3}=\lim _{h \rightarrow 0} \frac{9+h^{2}+6 h-3-h-6}{h}$                    $\left [ \because \left ( a+b \right )^{2}=\left ( a^{2}+b^{2}+2ab \right ) \right ]$

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{h^{2}+5 h}{h} \\\\ &=\lim _{h \rightarrow 0}(5+h)=5 \end{aligned}

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

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

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