Limit of a Rational Trigonometric Function $\lim_{x \to 0} \frac{\sin5x}{\sin4x}$ When solving a trigonometric limit such as:
$$\lim_{x \to 0} \frac{\sin(5x)}{\sin(4x)}$$
we rework the equation to an equivalent for to fit the limit of sine "rule":
$$\lim_{x \to 0}\frac{\sin(x)}{x}=1$$
so, we move forward in such a manner as follows:
$$=\lim_{x \to 0} \frac{\frac{5\sin(5x)}{5x}}{\frac{4\sin(4x)}{4x}}$$
$$=\frac{5}{4}\lim_{x \to 0} \frac{\frac{\sin(5x)}{5x}}{\frac{\sin(4x)}{4x}}$$
$$=\frac{5}{4}\cdot\frac{1}{1}$$
$$L=\frac{5}{4}$$
From that mindset, I am trying to find this trigonometric limit:
$$\lim_{x\to 2} \frac{\cos(x-2)-1}{x^{2}+x-6}$$
I know the Limit "rule" for cosine is:
$$\lim_{x\to 0} \frac{\cos(x)-1}{x}=0$$
If you use direct substitution in the original function you end up with an equation that is same in value to the rule presented. So, from that I am assuming that the answer is $0$. I am just trying to prove that in a step-by-step manner as I did with the sine limit.
P.S. I searched through many many pages of questions and didn't find something that helped. So, if I am repeating a question, I apologize that I missed it. 
Thanks for the help!

If you factor the denominator you would get:
$$\lim_{x\to 2} \frac{\cos(x-2)-1}{x^{2}+x-6}=\lim_{x\to 2} \frac{\cos(x-2)-1}{x-2}\frac{1}{x+3}=\ldots$$
