Explain solution RD Sharma class 12 chapter 8 Continuity exercise multiple choice question 16 maths

The correct option is (c)

Hint:

Use the given formula:

(i)  A function f(x) is said to be continuous at a point x = a of its domain, if

$\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)$

Given:

$f(x)=\frac{\sqrt{a^{2}-a x+x^{2}}-\sqrt{a^{2}+a x+x^{2}}}{\sqrt{a+x}-\sqrt{a-x}}$

Solution:

Using rationalization method,

$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\left(a^{2}-a x+x^{2}\right)-\left(a^{2}+a x+x^{2}\right)}{(\sqrt{a+x}-\sqrt{a-x})\left(\sqrt{a^{2}-a x+x^{2}}+\sqrt{a^{2}+a x+x^{2}}\right)}$

$\lim _{x \rightarrow 0} \frac{-2 a x(\sqrt{a+x}+\sqrt{a-x})}{2 x\left(\sqrt{a^{2}-a x+x^{2}}+\sqrt{a^{2}+a x+x^{2}}\right)}$

$\left.\lim _{x \rightarrow 0} \frac{-a(\sqrt{a+x}+\sqrt{a-x})}{\left(\sqrt{a^{2}-a x+x^{2}}+\sqrt{a^{2}+a x+x^{2}}\right.}\right)$

$=\lim _{x \rightarrow 0} \frac{-2 a(\sqrt{a})}{2 a}=-\sqrt{a}$

Understand that, f(x) is continuous for all x, then it is continuous at x = 0

\begin{aligned} &\lim _{x \rightarrow 0} f(x)=0 \\ &\lim _{x \rightarrow 0}-\sqrt{a}=f(0) \end{aligned}

So, the correct option is (c)