# Squeeze theorem and sequences

1. Oct 19, 2012

### kasperrepsak

1. The problem statement, all variables and given/known data

I have to proof that the sequence (2^n +n^2)/(3^n + 5n^4) converges en calculate its limit using the sqeeuze theorem.

2. Relevant equations
(2^n +n^2)/(3^n + 5n^4)

http://www.proofwiki.org/wiki/Squeeze_Theorem#Sequences

Theorem 1: Let p$\in$2N en x$\in$R with |x|< 1. Then the limit as n goes to infinity of (n^p)(x^n)=0

3. The attempt at a solution

I have noticed that if I divide the top and bottom by 3^n then i can use theorem 1 to calculate the limits of the top and the bottom.. the limit at the top will go to 0 and the bottom 1 .. so the limit of the whole goes to 0. My problem is that I dont know in what way i could use the squeeze theorem to proof this.

Last edited: Oct 19, 2012
2. Oct 19, 2012

### jbunniii

To use the squeeze theorem, try breaking the fraction into two parts:
$$\frac{2^n + n^2}{3^n + 5n^4} = \frac{2^n}{3^n + 5n^4} + \frac{n^2}{3^n + 5n^4}$$
Now try bounding each fraction on the right hand side by a simpler expression whose limit is easy to find.

3. Oct 19, 2012

### kasperrepsak

Something like this actually came into my head when i was pondering upon the question on the balcony.. : d If this is how long things will take on the exam. I am doomed. Thanks for the tip. I think i will manage to work this out.

4. Oct 19, 2012

### kasperrepsak

Actually, I cant see any way to solve this without using theorem 1. I cant find an upper bound for the two fractions.

5. Oct 19, 2012

### jbunniii

I'll show you how to find an upper bound for the first fraction. You can do something very similar for the second one. Notice that $5n^4 > 0$, so $3^n + 5n^4 > 3^n$. Therefore,

$$\frac{2^n}{3^n + 5n^4} < \frac{2^n}{3^n} = \left( \frac{2}{3}\right)^n$$
Also noting that the fraction is always positive, we have a lower bound of 0, so we may write
$$0 < \frac{2^n}{3^n + 5n^4} < \left( \frac{2}{3}\right)^n$$
The right-hand side goes to zero as $n \rightarrow \infty$, so the squeeze theorem applies.

6. Oct 19, 2012

### kasperrepsak

Thank you a lot, I see it now : )