Is it possible to find the limit of $\frac{\sin^2ax}{x\sqrt{1-\cos ax}}$?

Question

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?.

Answer 1

Since $$\lim_{x\rightarrow0}\frac{\sin{x}}{x}=1,$$ we obtain: $$\frac{\sin^2ax}{x\sqrt{1-\cos ax}}=\frac{\sin^2ax}{x\sqrt2\sin\frac{ax}{2}}\rightarrow\frac{a^2}{\sqrt2\cdot\frac{a}{2}}=a\sqrt2.$$

Answer 2

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$$LL=\lim_{x \to 0^-} f(x)=-a \sqrt{2}$$

Answer 3

Just expand the fraction with $\frac{\sqrt{1+\cos ax}}{\sqrt{1+\cos ax}}$.

So, you get for $a,x >0$

\begin{eqnarray*}\frac{\sin^2ax}{x\sqrt{1-\cos ax}} & = & \frac{\sin ax}{ax}\cdot \frac{\sin ax}{\sqrt{1-\cos^2 ax}} \cdot a \sqrt{1+\cos ax}\\ & \stackrel{x\to 0^+}{\longrightarrow} & a\sqrt 2 \end{eqnarray*}