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

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
