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$\lim_\limits{n\to\infty} \frac{2\cos(3n)+5\sin(n^2)}{n+1}$

$\lim_\limits{n\to\infty} \frac{(-1)^{n+1}+2^{-n}+\cos(n!)}{\sqrt n}$

I don't know how to solve this. I know I have to find at least one function to compare it with the ones I have but I can't think of one. I know the answers of the limits but I have to explain my answer using the Theorem. Can someone please help me?

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2 Answers 2

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HINT

For the first observe that

$$\frac{-7}{n+1}\le \frac{2\cos(3n)+5\sin(n^2)}{n+1}\le \frac{7}{n+1} $$

and for the second

$$\frac{-2}{\sqrt n}\le \frac{(-1)^{n+1} +2^{-n} + \cos(n!)}{\sqrt n}\le \frac{4}{\sqrt n}$$

now take the limit.

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  • $\begingroup$ But for which values $cos(3n)$ and $sin(n^2)$ are both $1$ or $-1$? Because that's when the numerator is $7$ or $-7$. $\endgroup$
    – Virginia Martín Herrera
    May 8, 2018 at 17:47
  • $\begingroup$ To apply the theorem we need just a lower and upper bound and $-7,+7$ is an estimation of those bounds. We could also use -100, and +150 it doesn't matter for the application of the squeeze theorem. The point is that the numerators in both cases are bounded. $\endgroup$
    – user
    May 8, 2018 at 17:52
  • $\begingroup$ @VirginiaMartínHerrera: That's irrelevant. The whole point is that, no matter what n is, you can say with certainty that the numerator (in the first one, for example) must be between -7 and 7. We don't care whether it's ever exactly -7 or 7, or what value of n would even make that happen. What matters is that, as n grows to infinity, the numerator is "stuck between two finite values" while the denominator grows and grows, resulting in a fraction that shrinks to 0. $\endgroup$
    – Brendan W. Sullivan
    May 8, 2018 at 17:52
  • $\begingroup$ Okay got it! thank you very much to both of you! $\endgroup$
    – Virginia Martín Herrera
    May 8, 2018 at 17:53
  • $\begingroup$ @VirginiaMartínHerrera You are welcome! Bye $\endgroup$
    – user
    May 8, 2018 at 18:04
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For both of these, all you really need is that sine and cosine are between -1 and 1. This lets you show that the numerators are bounded, so the ratio goes to zero.

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