calculus - find this limit without l'hopital's rule

I can't figure out how to get the limit in this problem. I know that ${1-\cos x \over x}=0$ but I'm not allowed to use L'Hopital's Rule. I also already know that the answer is $-{25 \over 36}$ but I don't know the steps in between. I've already tried multiplying by the conjugates of both the numerator and the denominator but neither are getting me anywhere close. Here is the question:

$$\lim_{x\to 0}{1-\cos 5x \over \cos 6x-1}$$

Answer

Outline: Our expression is equal to
$$-\frac{1+\cos 6x}{1+\cos 5x}\cdot\frac{1-\cos^25x}{1-\cos^2 6x},\tag{1}$$
which is
$$-\frac{1+\cos 6x}{1+\cos 5x}\cdot\frac{\sin^2 5x}{\sin^2 6x}.\tag{2}$$
To find
$$\lim_{x\to 0} \frac{\sin 5x}{\sin 6x},$$ rewrite as
$$\frac{5}{6}\lim_{x\to 0} \frac{\frac{\sin 5x}{5x}}{\frac{\sin 6x}{6x}}.$$