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We have that $\int_0^\infty \sin(r)r \, dr=\pm \infty$ (doesn't exist) so my guess is that

$\lim_{t\to \infty}\frac{\int_0^\infty \sin(r)r\,dr}{te^t}=\pm \infty$ even though the exponential grows faster.

We can also write this as $\lim_{t\to \infty}\lim_{R\to \infty}\frac{\int_0^R \sin(r)r\,dr}{te^t}$.

Thanks

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    $\begingroup$ Did you mean for the integral to have bounds which depend on $t$? $\endgroup$ – JimmyK4542 Jan 2 '15 at 2:15
  • $\begingroup$ no, I meant what I wrote. This is not a textbook question. $\endgroup$ – TKM Jan 2 '15 at 2:58
  • $\begingroup$ Perhaps they meant $\displaystyle\lim_{t\to\infty}\frac{\displaystyle\int_0^t\sin(r)~r~dr}{t~e^t}$ ? $\endgroup$ – Lucian Jan 2 '15 at 12:25
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Since $I=\int_0^{\infty}r\sin r\,dr$ doesn't exist, it follows that there is no real number $t$ for which $I/(te^t)$ exists, and a fortiori $\lim_{t\to\infty}(I/(te^t))$ can't possibly exist.

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  • $\begingroup$ Prettier and more concise,but basically the same answer I gave. $\endgroup$ – Mathemagician1234 Jan 2 '15 at 4:49
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The numerator is a DEFINITE integral,whicb is a number. So technically,only the denominator is a function of t. However,clearly the numerator goes to infinity,so it's not a limit that exists so by that reasoning alone, the limit doesn't exist. In that sense,the problem is really a trick question. You have to recognize that the numerator doesn't exist,which means the limit cannot exist! I made that mistake in my initial attempt to solve the problem-I attempted to use L'Hopital's rule-in essence, I treated the numerator as an indefinite integral Which is course,completely wrong. Which is why it really is critical to be absolutely clear what terms are present! Carelessness can cost!

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