$\lim_{x \to 0} \frac{\sin(a + b)x + \sin(a - b)x + sin(2ax)}{\cos^2 bx - cos^2 ax}$ Question : -

$$\lim_{x \to 0} \frac{\sin(a + b)x + \sin(a - b)x + \sin(2ax)}{\cos^2 bx - \cos^2 ax}$$

My attempt :-

$$\lim_{x \to 0} \frac{2* \sin ax * \cos bx + 2*\sin ax * \cos (ax)}{(\cos bx - \cos ax)*(\cos bx +\cos ax)}$$
  $$\lim_{x \to 0} \frac{2* \sin ax * (\cos bx + \cos (ax))}{(\cos bx - \cos ax)*(\cos bx +\cos ax)}$$
  $$\lim_{x \to 0} \frac{2* \sin ax}{(\cos bx - \cos ax)}$$
  $$\lim_{x \to 0} \frac{2* \sin ax}{(-2 * \sin {bx-ax\over2} * \sin {bx+ax\over2})}$$
  $$-\lim_{x \to 0} \frac{{\sin ax \over ax} * {ax }}{\frac{\sin {bx-ax\over2} * \sin {bx+ax\over2}}{ {bx+ax\over2} * {bx-ax\over2}} * {bx+ax\over2} * {bx-ax\over2}}$$
  $$-\lim_{x \to 0} \frac{ax}{{bx+ax\over2} * {bx-ax\over2}}$$
  $$-\lim_{x \to 0} \frac{ax}{{(bx)^2 - (ax)^2\over 4}}$$
  $$\lim_{x \to 0} \frac{4 * ax}{(ax)^2 -(bx)^2}$$
  $$\lim_{x \to 0} \frac{4 * a}{x (a^2 - b^2)}$$

There i hit the dead end, i can't do anything now.
Can anyone please tell me what i have done incorrect here ?
Sorry but i can't confirm the legitimacy of the question, i just found it on net and there is no answer given.  
 A: I'll assume that $a^2-b^2\ne0$ and $a\ne0$.
Notice that $\cos^2(bx)-\cos^2(ax)=(\cos(bx)+\cos(ax))(\cos(bx)-\cos(ax))$ and that the first factor has limit $2$, so it can be set away momentarily. Then
$$
\cos(bx)-\cos(ax)=1-\frac{b^2x^2}{2}+o(x^2)-1+\frac{a^2x^2}{2}+o(x^2)
=\frac{a^2-b^2}{2}x^2+o(x^2)
$$
and so the denominator becomes
$$
(a^2-b^2)x^2+o(x^2)
$$
because the other factor is $2+o(1)$.
The numerator is
$$
(a+b)x+(a-b)x+2ax+o(x)=4ax+o(x)
$$
and so the limit is infinite (different from the right than from the left).
Your computation seems correct, as you arrive essentially at the same point.

If $a=0$, the numerator becomes identically $0$, so the limit is $0$.
If $a^2=b^2$, then the denominator vanishes identically, so the limit is meaningless.
A: HINT:
$$\cos^2bx-\cos^2ax=\sin^2ax-\sin^2bx=\sin(a+b)x\sin(a-b)x$$ using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $
and we know $\lim_{h\to0}\dfrac{\sin h}h=1$
Clearly  the limit reduces to $$\dfrac{2a}{(a+b)(a-b)}\lim_{x\to0}\dfrac1x$$ which is non-existent
