# Proving Limit with Epsilon and Delta

1. Mar 27, 2014

### msell2

1. The problem statement, all variables and given/known data
Prove the following sequence {an} converges to L=1/2
an = n2/(2n2+n-1)

3. The attempt at a solution
Given ε>0 we can determine an N∈N so that |an - L|<ε for n≥N. We have:
|an-L|=|(n2/(2n2+n-1)-(1/2)|
= |(-n+1)/(2(2n-1)(n+1))|

I'm not sure what to do once I get to this point. Any help would be great!

2. Mar 27, 2014

### LCKurtz

If you change the numerator to (+n+1) you can get that fraction to be less or equal to$$\frac 1 {2(2n-1)}$$What happens if you drop the $-1$ too?

 Ignore the drop the -1 part. Maybe change the -1 to -n?

Last edited: Mar 27, 2014
3. Mar 27, 2014

### vela

Staff Emeritus
In a similar vein, you might want to show that $0<\frac{n-1}{2(2n-1)}<1$ for n>1.

4. Mar 28, 2014

### PeroK

The point behind these ideas is that it can be difficult to find N for a given ε unless the expression is quite simple, so you estimate the expression you have with something simpler.

E.g. suppose you wanted to show that 1/(n^2 + 6) -> 0. One way to do this is to note that:

0 < 1/(n^2+6) < 1/n

So, you can take N = 1/ε. Which is much simpler than trying to find a more precise N for the original sequence.

For your example, you want to find a simple expression that is greater than the one you have but still -> 0. That's the strategy, anyway. (See Vela's idea for how to do this.)

5. Mar 29, 2014

### msell2

Thanks for the help! Does this look right?

Given ε>0 we can determine an N$\in$N so that |an-L|<ε for n≥N. We have:
|an-L|=|$\frac{n^{2}}{2n^{2}+n-1}$-$\frac{1}{2}$|=|$\frac{2n^{2}}{2(2n^{2}+n-1}$-$\frac{2n^{2}+n-1}{2(2n^{2}+n-1}$|=|$\frac{-n+1}{2(2n^{2}+n-1}$|=|$\frac{-n+1}{2(2n-1)(n+1)}$|<ε.
Now take $\frac{n-1}{2(2n-1)}$. Since n-1>1 when n>1 and since 4n-2 > n-1 > 1 $\Rightarrow$ 0<$\frac{n-1}{2(2n-1)}$<1 when n>1. As long as we have n>1, we have that |$\frac{-n+1}{2(2n-1)(n+1)}$|<$\frac{n-1}{2(2n-1)}$. We want to bound this by ε>0. Take ε>0. If we choose N to be the smallest integer greater than or equal to max{1,1/ε} then we have that, for n≥N:
|$\frac{n^{2}}{2n^{2}+n-1}$-$\frac{1}{2}$|<$\frac{1}{n}$<$\frac{1}{N}$≤ε
where the last line follows directly if N≥1/ε and is also satisfied for N=1>1/ε since this implies 1/N<ε.

6. Mar 29, 2014

### vela

Staff Emeritus
Is that last inequality what you intended?

7. Mar 29, 2014

### msell2

What is wrong with that inequality?

8. Mar 30, 2014

### vela

Staff Emeritus
How are you going from
to

9. Mar 31, 2014

### dirk_mec1

n-1< 4n -2 => 1<3n which is true for n>1 get it?