# 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?

• OMG -- please tell me you didn't do this: $$\frac{\sin (x-5)}{x^2 - 2x - 15} = \require{cancel}\frac{\sin \cancel{(x-5)}}{(x+3)\cancel{(x-5)}} = \frac{\sin}{(x+3)}$$ – MPW Oct 10 '14 at 20:53
• Perhaps this is the limit as $X$ approaches $5$: $$5 X$$ $$5 \! X$$ $$5 \!\! X$$ $$5 \!\!\! X$$ $$5 \!\!\!\! X$$ Sorry, couldn't resist. It's Friday afternoon. – MPW Oct 10 '14 at 21:03
• @OP - You can't cancel the $(x - 5)$ in $\frac{sin(x - 5)}{(x + 3)(x - 5)}$ $(x - 5)$ is the argument to the $sin$ function, not a free-floating expression. – Tyler Gaona Oct 10 '14 at 23:00
• @TylerGaona @OP doesn't work... the OP is always notified whenever someone comments its post. – Braiam Oct 10 '14 at 23:36
• $\sin$ is a function, it isn't a number. This function takes a number $x$ and outputs a number denoted as $\sin(x)$. So the notation $\sin(\text{number})$ doesn't mean that you take $\sin$ and multiply it by that number, that's simply meaningless. – Hakim Oct 11 '14 at 12:01

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!!

• Sorry if I did not know I can't. Can I know why? So i don't make this mistake again – Simon Oct 10 '14 at 21:06
• $(x-5)$ is the argument of the sine, it's not $\sin\mathbf{\times}(x-5)$. Therefore you can't cancel it. – rae306 Oct 10 '14 at 21:14
• @Kabama : $\:$ It's like canceling the 3s to go from $\frac{13}{23}$ to $\frac12$. $\;\;\;\;$ – user57159 Oct 11 '14 at 5:45

\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 $x-5=y$)

$\sin(x−5)$: 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):

$$\lim_{x\to 5}\frac{x-5}{(x+3)(x-5)}$$

$$\frac{\sin(x-5)}{x^2-2x-15}=\frac{\sin(x-5)}{x^2-2x+1-16}=\frac{\sin(x-5)}{(x-1)^2-4^2}=$$ $$=\frac{\sin(x-5)}{(x-1-4)(x-1+4)}=\frac{\sin(x-5)}{(x-5)}\frac{1}{(x+3)}\to\frac{1}{8}\text{if}x\to5$$

• Seems a little short on explanation, and it's not a hint so much as a solution. – Brilliand Oct 10 '14 at 22:44