#### Need solution for RD Sharma Maths Class 12 Chapter 8 Continuity Excercise 8.1 Question 3

$x=3$

Hint:

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

Solution:

Given,

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

We observe,

[LHL at $x=3$ ]

\begin{aligned} &\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} \end{aligned}

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

\begin{aligned} &=\lim _{h \rightarrow 0}(6-h) \\ &=6 \end{aligned}

[RHL at $x=3$ ]

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

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

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

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{h(h+6)}{h} \\\\ &=\lim _{h \rightarrow 0}(6+h) \\\\ &=6 \end{aligned}

Given

\begin{aligned} &f(3)=6 \\ \\&\therefore \lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{-}} f(x)=f(3) \end{aligned}

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