$f(x) = \dfrac{\sin(x-5)}{x^2-2x-15}$
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?
@OP
doesn't work... the OP is always notified whenever someone comments its post. $\endgroup$ – Braiam Oct 10 '14 at 23:36