2
$\begingroup$

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

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

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?

$\endgroup$
0
3
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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 '18 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 '18 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 '18 at 17:52
  • $\begingroup$ Okay got it! thank you very much to both of you! $\endgroup$ – Virginia Martín Herrera May 8 '18 at 17:53
  • $\begingroup$ @VirginiaMartínHerrera You are welcome! Bye $\endgroup$ – user May 8 '18 at 18:04
1
$\begingroup$

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.

$\endgroup$

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.