# Use the Squeeze (Sandwich) Theorem to solve this limits:

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

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

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?

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.

• But for which values $cos(3n)$ and $sin(n^2)$ are both $1$ or $-1$? Because that's when the numerator is $7$ or $-7$. – Virginia Martín Herrera May 8 '18 at 17:47
• 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. – user May 8 '18 at 17:52
• @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. – Brendan W. Sullivan May 8 '18 at 17:52
• Okay got it! thank you very much to both of you! – Virginia Martín Herrera May 8 '18 at 17:53
• @VirginiaMartínHerrera You are welcome! Bye – user May 8 '18 at 18:04

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.