How to find $\lim_{x \to 0}\frac{\cos(ax)-\cos(bx) \cos(cx)}{\sin(bx) \sin(cx)}$ How to find  $$\lim_\limits{x \to 0}\frac{\cos (ax)-\cos (bx) \cos(cx)}{\sin(bx) \sin(cx)}$$ 
I tried using L Hospital's rule but its not working!Help please!

Before using L'Hospital, turn the products to sums
$$\frac{\cos(ax)-\cos(bx)\cos(cx)}{\sin(bx)\sin(cx)}=\frac{2\cos(ax)-\cos((b-c)x)-\cos((b+c)x)}{\cos((b-c)x)-\cos((b+c)x)}.$$
Then by repeated application
$$\frac{2a\sin(ax)-(b-c)\sin((b-c)x)-(b+c)\sin((b+c)x)}{(b-c)\sin((b-c)x)-(b+c)\sin((b+c)x)},$$
and
$$\frac{2a^2\cos(ax)-(b-c)^2\cos((b-c)x)-(b+c)^2\cos((b+c)x)}{(b-c)^2\cos((b-c)x)-(b+c)^2\cos((b+c)x)}.$$
The limit is
$$\frac{2a^2-(b-c)^2-(b+c)^2}{(b-c)^2-(b+c)^2}=-\frac{a^2-b^2-c^2}{2bc}.$$
