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:
$$\lim_{x\to 0}{1-\cos 5x \over \cos 6x-1}$$
 A: 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
$$\frac{5}{6}\lim_{x\to 0} \frac{\frac{\sin 5x}{5x}}{\frac{\sin 6x}{6x}}.$$
A: Hint.  First, note that
$$\lim_{x\to0}\frac{\sin ax}{x}=a\ ,$$
which can be proved by geometric arguments without using L'Hopital's rule.
We have
$$\eqalign{\frac{1-\cos ax}{x^2}
  &=\frac{1-\cos ax}{x^2}\frac{1+\cos ax}{1+\cos ax}\cr
  &=\Bigl(\frac{\sin ax}{x}\Bigr)^2\frac{1}{1+\cos ax}\ .\cr}$$
You should now be able to find the limit of this expression as $x\to0$, and hence solve your problem.
A: You could also use Taylor series since $$\cos(y)=1-\frac{y^2}{2}+O\left(y^3\right)$$ Now replace successively $y$ by $5x$ and get the numerator $$1-\cos(5x)=\frac{25 x^2}{2}+O\left(x^3\right)$$ and get the denominator replacing $y$ by $6x$, so $$\cos(6x)-1=-18 x^2+O\left(x^3\right)$$ So the ratio is what you said.
If you use more terms for each development and perform long division, you should end with $${1-\cos (5x) \over \cos (6x)-1}=-\frac{25}{36}-\frac{275 x^2}{432}-\frac{1595 x^4}{2592}+O\left(x^5\right)$$ which shows how the limit is approached when $x$ goes to $0$.
