Evaluating $\lim_{x\rightarrow 0}\frac{\cos(ax) - \cos(bx)}{\sin^2(x)}$ The problem is, as stated:

$$\lim_{x\rightarrow 0}\frac{\cos(ax) - \cos(bx)}{\sin^2(x)}$$
  Of course, a and b are real numbers.

I tried implementing the trig identity:
$$\cos(ax)-\cos(bx) = -2\sin\frac{(ax-bx)}{2}\sin\frac{(ax+bx)}{2}$$
But that didn't really take me anywhere. 
In my book, similar problems were often solved this way but here that doesn't seem to work.
One obviously has to try and use the known limit to reduce this to a more easy limit:
$$\lim_{x \rightarrow 0}\frac{\sin(x)}{x} = 1$$
Any help would be much appreciated.
 A: The identity given in your question is useful for evaluating the limit: as $x\to 0$ we have
$$\frac{\cos(ax) - \cos(bx)}{\sin^2(x)} = -2\frac{\frac{\sin\frac{(ax-bx)}{2}}{x}\cdot \frac{\sin\frac{(ax+bx)}{2}}{x}}{\frac{\sin x}{x}\cdot\frac{\sin x}{x}}\to -2\frac{\frac{a-b}{2}\cdot \frac{a+b}{2}}{1\cdot 1}=\frac{b^2-a^2}{2}$$
where we used the fact that for any real $\alpha$,
$$\lim_{x\to 0}\frac{\sin(\alpha x)}{x}=\alpha\lim_{x\to 0}\frac{\sin(\alpha x)}{(\alpha x)}=\alpha\cdot 1=\alpha.$$
A: \begin{align}\lim_{x\to0}\frac{\cos(ax)-\cos(bx)}{\sin^2x}&=\lim_{x\to0}\frac{x^2}{\sin^2x}\cdot\frac{\cos(ax)-\cos(bx)}{x^2}\\&=\lim_{x\to0}\frac{\cos(ax)-\cos(bx)}{x^2}\\&=\frac{b^2-a^2}2\end{align}since$$\cos(ax)-\cos(bx)=-\frac12(a^2-b^2)x^2+o(x^2).$$
A: $$\lim_{x\rightarrow 0}\frac{\cos(ax) - \cos(bx)}{\sin^2(x)}=\lim_{x\rightarrow 0}\frac{x^2}{\sin^2(x)}\lim_{x\rightarrow 0}\frac{\cos(ax) - \cos(bx)}{x^2}\\=-\lim_{x\rightarrow 0}\frac{1- \cos(ax)}{x^2}+\lim_{x\rightarrow 0}\frac{1- \cos(bx)}{x^2}=\frac{ b^2-a^2}{2}$$
Given that $$\lim_{x\rightarrow 0}\frac{1- \cos(x)}{x^2}=1/2$$
$$\lim_{x\rightarrow 0}\frac{ \sin(x)}{x}=1. $$
A: Hint For Finishing The Approach In The Question
$$
\frac{\cos(ax)-\cos(bx)}{\sin^2(x)}=2\frac{\sin\left(\frac{b-a}2\,x\right)}{\sin(x)}\frac{\sin\left(\frac{b+a}2\,x\right)}{\sin(x)}
$$

Hint For Another Approach
$$
\begin{align}
\frac{\cos(ax)-\cos(bx)}{\sin^2(x)}
&=\frac{1-\cos(bx)}{\sin^2(x)}-\frac{1-\cos(ax)}{\sin^2(x)}\\
&=\frac1{1+\cos(bx)}\frac{\sin^2(bx)}{\sin^2(x)}-\frac1{1+\cos(ax)}\frac{\sin^2(ax)}{\sin^2(x)}
\end{align}
$$
A: You are on the right track,
$$ \lim_{x\rightarrow 0} \frac{\cos (ax)-\cos (bx)}{\sin^2 x}=\lim_{x\rightarrow0}\frac{-2 \sin \left(\frac{a+b}{2}x \right)\sin \left(\frac{a-b}{2}x \right)}{\sin^2 x}=\lim_{x\rightarrow0}-2\frac{\frac{\sin \left(\frac{a+b}{2}x \right)}{x}}{\frac{\sin x}{x}}\frac{\frac{\sin \left(\frac{a-b}{2}x \right)}{x}}{\frac{\sin x}{x}}.$$
It should be obvious now.
A: You can use L' Hospital rule
$$\lim_{x\rightarrow 0}\frac{\cos(ax) - \cos(bx)}{\sin^2(x)}$$
$$=\lim_{x\rightarrow 0}\frac{-a\sin(ax) +b \sin(bx)}{2\sin(x)\cos(x)}$$
$$=\lim_{x\rightarrow 0}\frac{-a^2\cos(ax) +b^2 \cos(bx)}{2\cos^2(x)-2\sin^2(x)}=\frac{b^2-a^2}{2}$$
Where we applied L'Hospital's rule twice
A: i would write
$$\frac{-2\left(\sin x\left(\frac{a-b}{2}\right)\right)\sin\left(x\left(\frac{a+b}{2}\right)\right)x^2\left(\frac{a^2-b^2}{4}\right)}{x\left(\frac{a-b}{2}\right) x\left(\frac{a+b}{2}\right)\sin^2(x)}$$
A: I believe this approach is simpler: since $1-\cos(cx)=2\sin^2\frac{cx}{2}$ and $\lim_{x\to 0}\frac{\sin(dx)}{\sin(x)}=d$,
$$ \lim_{x\to 0}\frac{\cos(ax)-\cos(bx)}{\sin^2(x)}=\lim_{x\to 0}\frac{2\sin^2\frac{bx}{2}-2\sin^2\frac{ax}{2}}{\sin^2(x)}=2\left[\left(\frac{b}{2}\right)^2-\left(\frac{a}{2}\right)^2\right]=\color{blue}{\frac{b^2-a^2}{2}}.$$
A: From the Taylor series of $\sin(x),\cos(x)$ and using $O$-notation, we know that, for $x\to 0$,
\begin{eqnarray*}
\sin x &=&x+O\left( x^{3}\right)  \\
\cos x &=&1-\frac{1}{2}x^{2}+O\left( x^{3}\right). 
\end{eqnarray*}
Then 
$$
\sin ^{2}x=x^{2}+O\left( x^{4}\right) 
$$
and 
$$
\frac{\cos ax-\cos bx}{\sin ^{2}x}=\frac{\left( -\frac{1}{2}a^{2}+\frac{1}{2}
b^{2}\right) x^{2}+O\left( x^{3}\right) }{x^{2}+O\left( x^{4}\right) }.
$$
Hence
$$
\lim_{x\rightarrow 0}\frac{\cos ax-\cos bx}{\sin ^{2}x}=\lim_{x\rightarrow 0}
\frac{x^{2}\left( -\frac{1}{2}a^{2}+\frac{1}{2}b^{2}\right) }{x^{2}}=\frac{1
}{2}b^{2}-\frac{1}{2}a^{2}.
$$
