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?
 A: $$\begin{align}\lim_{x\to 5}\frac{\sin(x-5)}{x^2-2x-15}&=\lim_{x\to 5}\frac{\sin(x-5)}{(x-5)(x+3)}\\&=\lim_{x\to 5}\frac{\sin(x-5)}{x-5}\cdot\frac{1}{x+3}\\&=1\cdot \frac{1}{5+3}\\&=\frac{1}{8}.\end{align}$$
Here, note that 
$$\lim_{y\to 0}\frac{\sin y}{y}=1.$$
(set $x-5=y$)
A: Using L'Hôpital's Rule:
$$\lim_{x\to5}\frac{\sin(x-5)}{x^2-2x-15}=\lim_{x\to5}\frac{\cos(x-5)}{2x-2}=\frac{1}{8}$$

Edit:
Your '$\dfrac{\sin(x-5)}{x^2-2x-15}=\dfrac{\sin}{ ( x+3)}$' is one of the most deadly sins of mathematics! One cannot cancel here!!
A: $\sin(x−5)$: Hm… $x$ is going to 5? How does $\sin(x)$ behave as ${x\to 0}$? It behaves as if $\sin(x)$ was $x$ (because $\sin(x) \approx x$ near $x=0$ and the derivative of $\sin(x)$ (which is $\cos(x)$) is the derivative of $x$ when both are evaluated at 0).
So since $\sin(x)$ is basically an identity function in this limit, you have (after factoring the denominator):
$$\lim_{x\to 5}\frac{x-5}{(x+3)(x-5)}$$
A: $$\frac{\sin(x-5)}{x^2-2x-15}=\frac{\sin(x-5)}{x^2-2x+1-16}=\frac{\sin(x-5)}{(x-1)^2-4^2}=$$
$$=\frac{\sin(x-5)}{(x-1-4)(x-1+4)}=\frac{\sin(x-5)}{(x-5)}\frac{1}{(x+3)}\to\frac{1}{8}\text{if}x\to5$$
