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!

One can use only the following standard limits (without L'Hospital's rule nor Taylor series)
\begin{equation*}
\lim_{u\rightarrow 0}\frac{1-\cos u}{u^{2}}=\frac{1}{2}\ \ \ \ \ \ \ \ ,\
\lim_{u\rightarrow 0}\cos u=\cos 0=1\ \ \ \ \ \ \ \ \lim_{u\rightarrow 0}%
\frac{\sin u}{u}=1
\end{equation*}
To this end, transform the original expression as follows
\begin{equation*}
\frac{\cos ax-\cos bx\cos cx}{\sin bx\sin cx}=\frac{\cos ax-\cos bx\cos cx}{%
x^{2}}\frac{bx}{\sin bx}\frac{cx}{\sin cx}\frac{1}{bc}
\end{equation*}
Note that 
\begin{equation*}
\cos bx\cos cx=(1-\cos bx)(1-\cos cx)+\cos cx+\cos bx-1
\end{equation*}
then
\begin{eqnarray*}
\frac{\cos ax-\cos bx\cos cx}{\sin bx\sin cx} &=&\left( \frac{\cos
ax-(1-\cos bx)(1-\cos cx)-\cos cx-\cos bx+1}{x^{2}}\right)  \times \frac{bx}{\sin bx}\frac{cx}{\sin cx}\frac{1}{bc} \\
&=&\left( \frac{\cos ax-1}{(ax)^{2}}a^{2}+\frac{1-\cos bx}{(bx)^{2}}b^{2}+%
\frac{1-\cos cx}{(cx)^{2}}c^{2}+\frac{1-\cos cx}{(cx)^{2}}(\cos
bx-1)c^{2}\right)  \times \frac{bx}{\sin bx} \frac{cx}{\sin cx} \frac{1}{bc}.
\end{eqnarray*}
Passing to the limit as $x$ tends to $0$ and using standard limits cited
above one gets
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\cos ax-\cos bx\cos cx}{\sin bx\sin cx} &=&\left( 
\frac{-1}{2}a^{2}+\frac{1}{2}b^{2}+\frac{1}{2}c^{2}+\frac{1}{2}%
(1-1)c^{2}\right) \times 1\cdot 1\cdot \frac{1}{bc} \\
&=&\frac{-a^{2}+b^{2}+c^{2}}{bc}.
\end{eqnarray*}
