<s>[INST] Find the limit as $x$ approaches $5$ $f(x)  = \dfrac{\sin(x-5)}{x^2-2x-15}$
Find the limit as $x$ approaches $5$.
I got up to : $\dfrac{\sin}{ ( x+3)}$.
I know the answer is $\frac18$ but I just don't know how to get it.
Unfortunately, I did cancel out the (x-5) =(. Is it because the the numerator (x-5) is considered an angle? like sin theta? and is not similar to the one in the denominator?
 [/INST] $$\frac{\sin(x-5)}{x^2-2x-15}=\frac{\sin(x-5)}{x^2-2x+1-16}=\frac{\sin(x-5)}{(x-1)^2-4^2}=$$
$$=\frac{\sin(x-5)}{(x-1-4)(x-1+4)}=\frac{\sin(x-5)}{(x-5)}\frac{1}{(x+3)}\to\frac{1}{8}\text{if}x\to5$$
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