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?