<s>[INST] Is it possible to find the limit of $\frac{\sin^2ax}{x\sqrt{1-\cos ax}}$? The problem is as follows:

A certain tv signal is modeled by the function shown below:
$f(x)=\frac{\sin^2ax}{x\sqrt{1-\cos ax}}$
where $a>0$
Find the $\lim_{x\rightarrow 0^+} f(x)$.

The alternatives given in this problem are as follows:
$\begin{array}{ll}
1.&-a\sqrt{2}\\
2.&\frac{\sqrt{a}}{2}\\
3.&\sqrt{2}\sqrt{a}\\
4.&\sqrt{2}\\
5.&-\sqrt{a}\\
\end{array}$
How exactly should I assess this problem?.
I'm confused about the simbol used in the limit but I think the intended meaning is to find the limit of the function where $x$ approaches to positive?.
Attempting to insert the zero in the function as it is given would yield an infinite value. Thus I thought to reduce the trigonometric expression by doing this:
$f(x)=\frac{\sin^2ax}{x\sqrt{1-\cos ax}}$
$\frac{1-\cos^2 ax}{x\sqrt{1-\cos ax}}\times\frac{\sqrt{1-\cos ax}}{\sqrt{1-\cos ax}}$
$\frac{(1-\cos ax)(1+\cos ax )(\sqrt{1-\cos ax})}{x(1-\cos ax)}$
Simplifying terms in both denominator and numerator it yields
$\frac{(1+\cos ax )(\sqrt{1-\cos ax})}{x}$
By inserting the expression in the numerator inside the square root I'm getting:
$\frac{\sqrt{(1+\cos ax)^2(1-\cos ax})}{x}$
Expanding the whole expression I'm getting:
$\frac{\sqrt{(1^2+2\cos ax+\cos^2ax)(1-\cos ax})}{x}$
$\frac{\sqrt{1+2\cos ax+\cos^2ax-\cos ax-2\cos^2ax-\cos^3 ax}}{x}$
$\frac{\sqrt{1+\cos ax-\cos^2ax-\cos^3 ax}}{x}$
and that's how far I went. What exactly should be done here?. Can someone help me here?.
 [/INST] Note that $\sqrt{g^2(x)}=|g(x)|$
$$L=\lim_{x\to 0} \frac{\sin^2 ax}{x\sqrt{1-\cos ax}}= \lim_{x\to 0} \frac{\sin^2 ax}{x \sqrt{2}~|\sin (ax/2)|} =\lim_{x\to 0} \frac{a^2x^2}{x|ax/2|\sqrt{2}}$$
So $$RL=\lim_{x \to 0^+} f(x)=a\sqrt{2}$$
and$$LL=\lim_{x \to 0^-} f(x)=-a \sqrt{2}$$
</s>