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!
 A: i will expand on the hint given by dr. mv.  we will compute the numerator and denominators separately. we have 
$$\begin{align}\cos ax - \cos bx \cos cx &= 1 - a^2x^2/2 + \cdots - (1 - b^2x^2/2+\cdots)(1-c^2x^2/2+\cdots)\\&= \frac12 (b^2 + c^2 - a^2) x^2 + \cdots\\
\sin bx \sin cx &= (bx + \cdots)(cx + \cdots) = bcx^2 + \cdots\end{align}$$
therefore, $$\lim_{x \to 0}\frac{\cos ax - \cos bx \cos cx}{\sin bx \sin cx} = \frac{b^2 + c^2 - a^2}{2bc}.$$
A: HINT:
$$\cos ax =1-\frac12 (ax)^2+O(x^4)$$
and
$$\sin ax= ax+O(x^3)$$
A: $$\cos(ax)-\cos(bx)\cos(cx)=\dfrac{2\cos(ax)-2\cos(bx)\cos(cx)}2$$
$$=\dfrac{2\cos(ax)-[\cos(b-c)x+\cos(b+c)x}2$$
$$=\dfrac{1-\cos(b-c)x+1-\cos(b+c)x-2[1-\cos(ax)]}2$$
Now $\dfrac{1-\cos px}{\sin qx\sin rx}=\dfrac{1-\cos^2px}{(1+\cos px)\sin qx\sin rx}$
$=p^2\left(\dfrac{\sin px}{px}\right)^2\cdot\dfrac1{q\cdot\dfrac{\sin qx}{qx}}\cdot\dfrac1{r\cdot\dfrac{\sin rx}{rx}}\cdot\dfrac1{(1+\cos px)}$
A: Using the fact that
$ \lim_{x \to 0} \frac{\sin (\alpha x)}{x} = \alpha $
you may write
$$ \lim_{x \to 0} \frac{ \cos(ax) - \cos(bx) \cos(cx) }{ \sin(bx) \sin(cx) }
$$
$$
= \lim_{x \to 0} \frac{ \cos(ax) - \cos(bx)\cos(cx)}{\sin(bx)\sin(cx)} \frac{(bx)(cx)}{(bx)(cx)}
$$
$$
= \lim_{x \to 0} \frac{\cos(ax) - \cos(bx)\cos(cx)}{bcx^2}
$$
Then you may apply L'Hospital's rule here.
A: 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
$$\frac{2a^2-(b-c)^2-(b+c)^2}{(b-c)^2-(b+c)^2}=-\frac{a^2-b^2-c^2}{2bc}.$$
A: 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*}
