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!

Solutions Collecting From Web of "Limit of a Rational Trigonometric Function $\lim_{x \to 0} \frac{\sin5x}{\sin4x}$"

If you factor the denominator you would get:

$$\lim_{x\to 2} \frac{\cos(x-2)-1}{x^{2}+x-6}=\lim_{x\to 2} \frac{\cos(x-2)-1}{x-2}\frac{1}{x+3}=\ldots$$

For an easy evaluation you can use La’Hospital’s rule.
$$\lim_{x\to 2}\frac{\cos(x-2)-1}{x^2+x-6} = \lim_{x\to 2}\frac{-\sin(x-2)}{2x+1}=0$$