#### Need solution for RD Sharma Maths Class 12 Chapter 8 Continuity Excercise 8.1 Question 4

Continuous

Hint:

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

Solution:

Given,

$f(x)=\left(\begin{array}{cc} \frac{x^{2}-1}{x-1} & , \text { if } x \neq 1 \\\\ 2 & , \text { if } x=1 \end{array}\right)$

We observe,

[LHL at $x=1$ ]

\begin{aligned} &\lim _{h \rightarrow 0} \frac{(1-h)^{2}-1}{(1-h)-1} \\\\ &=\lim _{h \rightarrow 0} \frac{1+h^{2}-2 h-1}{1-h-1} \end{aligned}

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

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

[RHL at $x=1$]

\begin{aligned} &\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h) \\\\ &\lim _{h \rightarrow 0} \frac{(1+h)^{2}-1}{(1+h)-1} \end{aligned}

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

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

Given

\begin{aligned} &f(1)=2 \\\\ &\therefore \lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{-}} f(x)=f(1) \end{aligned}

Hence, $f\left ( x \right )$ is continuous at $x=1$.