[Math] How to find $\lim_{x \to 0}\frac{\cos(ax)-\cos(bx) \cos(cx)}{\sin(bx) \sin(cx)}$

Question

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!

Answer 1

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, $$\lim_{x \to 0}\frac{\cos ax - \cos bx \cos cx}{\sin bx \sin cx} = \frac{b^2 + c^2 - a^2}{2bc}.$$