1
$\begingroup$

$f(x) = \dfrac{\sin(x-5)}{x^2-2x-15}$

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?

$\endgroup$
  • 23
    $\begingroup$ OMG -- please tell me you didn't do this: $$\frac{\sin (x-5)}{x^2 - 2x - 15} = \require{cancel}\frac{\sin \cancel{(x-5)}}{(x+3)\cancel{(x-5)}} = \frac{\sin}{(x+3)}$$ $\endgroup$ – MPW Oct 10 '14 at 20:53
  • 20
    $\begingroup$ 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. $\endgroup$ – MPW Oct 10 '14 at 21:03
  • 3
    $\begingroup$ @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. $\endgroup$ – Tyler Gaona Oct 10 '14 at 23:00
  • 4
    $\begingroup$ @TylerGaona @OP doesn't work... the OP is always notified whenever someone comments its post. $\endgroup$ – Braiam Oct 10 '14 at 23:36
  • 2
    $\begingroup$ $\sin$ 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. $\endgroup$ – Hakim Oct 11 '14 at 12:01
10
$\begingroup$

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!!

$\endgroup$
  • $\begingroup$ Sorry if I did not know I can't. Can I know why? So i don't make this mistake again $\endgroup$ – Simon Oct 10 '14 at 21:06
  • 5
    $\begingroup$ $(x-5)$ is the argument of the sine, it's not $\sin\mathbf{\times}(x-5)$. Therefore you can't cancel it. $\endgroup$ – rae306 Oct 10 '14 at 21:14
  • 2
    $\begingroup$ @Kabama : $\:$ It's like canceling the 3s to go from $\frac{13}{23}$ to $\frac12$. $\;\;\;\;$ $\endgroup$ – user57159 Oct 11 '14 at 5:45
6
$\begingroup$

$$\begin{align}\lim_{x\to 5}\frac{\sin(x-5)}{x^2-2x-15}&=\lim_{x\to 5}\frac{\sin(x-5)}{(x-5)(x+3)}\\&=\lim_{x\to 5}\frac{\sin(x-5)}{x-5}\cdot\frac{1}{x+3}\\&=1\cdot \frac{1}{5+3}\\&=\frac{1}{8}.\end{align}$$ Here, note that $$\lim_{y\to 0}\frac{\sin y}{y}=1.$$ (set $x-5=y$)

$\endgroup$
1
$\begingroup$

$\sin(x−5)$: 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):

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ Seems a little short on explanation, and it's not a hint so much as a solution. $\endgroup$ – Brilliand Oct 10 '14 at 22:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.