#### provide solution for RD Sharma maths class 12 chapter 8 Continuity exercise  Fill in the blanks question 2

Answer: $\pm 1$

Hint:

Use the formula of  $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$  to solve $\frac{\sin ^{2} a x}{x^{2}}$

Given:

$f(x)=\left\{\begin{array}{cl} \frac{\sin ^{2} a x}{x^{2}}, & x \neq 0 \\ 1, & x=0 \end{array}\right.$  is continuous at $x=0$

Solution:

If $f(x)$ is continuous at $x=0$, then $f(x)=\frac{\sin ^{2} a x}{x^{2}}$ and $g(x)=1$ are equal.

LHL,

\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{\sin ^{2} a x}{x^{2}} \\ &=\lim _{x \rightarrow 0^{-}} \frac{\sin ^{2} a x}{a^{2} x^{2}}\left(a^{2}\right) \\ \end{aligned}

$=\lim_{x\rightarrow 0^{-}}\left ( \frac{\sin ax}{ax} \right )^{2}\left ( a^{2} \right )$

$=a^{2}$                                                                $\left[\because \frac{\sin ^{2} a x}{a^{2} x^{2}}=1\right]$

RHL,

$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} 1=1$

As $f(x)$ is continuous at $x=0$, LHL =RHL

\begin{aligned} &a^{2}=1 \\ &a=\pm 1 \end{aligned}