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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!

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up vote 3 down vote accepted

If you factor the denominator you would get:

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

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wow. I feel like a moron now. apparently i need a break if I didn't see that. Thanks! I'll accept your answer as soon as it allows. – PathemaMike Jul 2 '14 at 0:07
@PathemaMike When you plug in $x=2$ and you find that the denominator is equal to $0$, you'll immediately understand that $2$ is a root of $x^2+x-6$, so we can write it as $(x-2)(x-\alpha)$. And we note that we have $\cos(x-2) $, so we could write it as our general limit. Hope this clarifies things for you. ;-) – Hakim Jul 2 '14 at 0:10
Thanks! I think I just need to step away for a minute, and let my mind do other things if I didn't see such an obvious solution! – PathemaMike Jul 2 '14 at 0:16
@PathemaMike You're welcome! $\overset{\cdot\cdot}\smile$ – Hakim Jul 2 '14 at 0:39

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