Need solution for RD Sharma Maths Class 12 Chapter 8 Continuity Excercise 8.1 Question 10 Subquestion (vi)

Continuous at  $x=1$

Hint:

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

Solution:

Given,

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

We observe

[LHL at $x=1$ ]

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

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

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

RHL=LHL

Therefore, $f\left ( x \right )$ is continuos at  $x=1$