Using the Squeeze Theorem to Prove Convergence of a Sequence

[kasperrepsak]

Homework Statement

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

Homework Equations

(2^n +n^2)/(3^n + 5n^4)

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

Theorem 1: Let p\in2N en x\inR with |x|< 1. Then the limit as n goes to infinity of (n^p)(x^n)=0

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 don't know in what way i could use the squeeze theorem to proof this.

 

[jbunniii]

Homework Statement

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

Homework Equations

(2^n +n^2)/(3^n + 5n^4)

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

Theorem 1: Let p\in2N en x\inR with |x|< 1. Then the limit as n goes to infinity of (n^p)(x^n)=0

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 don't know in what way i could use the squeeze theorem to proof this.

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.

 

[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.

 

[kasperrepsak]

Actually, I can't see any way to solve this without using theorem 1. I can't find an upper bound for the two fractions.

 

[jbunniii]

Actually, I can't see any way to solve this without using theorem 1. I can't find an upper bound for the two fractions.

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 &gt; 0, so 3^n + 5n^4 &gt; 3^n. Therefore,

\frac{2^n}{3^n + 5n^4} &lt; \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 &lt; \frac{2^n}{3^n + 5n^4} &lt; \left( \frac{2}{3}\right)^n
The right-hand side goes to zero as n \rightarrow \infty, so the squeeze theorem applies.

 

[kasperrepsak]

Thank you a lot, I see it now : )