Proving Limit with Epsilon and Delta

In summary, we begin by using the definition of convergence to show that the sequence {an} converges to L=1/2. We then manipulate the expression for |an-L| to find a simpler expression that is still less than ε. By choosing N to be the smallest integer greater than or equal to max{1,1/ε}, we can show that the sequence converges to L=1/2.
  • #1
msell2
15
0

Homework Statement


Prove the following sequence {an} converges to L=1/2
an = n2/(2n2+n-1)

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!
 
Physics news on Phys.org
  • #2
msell2 said:

Homework Statement


Prove the following sequence {an} converges to L=1/2
an = n2/(2n2+n-1)

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!

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?

[Edit] Ignore the drop the -1 part. Maybe change the -1 to -n?
 
Last edited:
  • #3
In a similar vein, you might want to show that ##0<\frac{n-1}{2(2n-1)}<1## for n>1.
 
  • #4
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
Thanks for the help! Does this look right?

Given ε>0 we can determine an N[itex]\in[/itex]N so that |an-L|<ε for n≥N. We have:
|an-L|=|[itex]\frac{n^{2}}{2n^{2}+n-1}[/itex]-[itex]\frac{1}{2}[/itex]|=|[itex]\frac{2n^{2}}{2(2n^{2}+n-1}[/itex]-[itex]\frac{2n^{2}+n-1}{2(2n^{2}+n-1}[/itex]|=|[itex]\frac{-n+1}{2(2n^{2}+n-1}[/itex]|=|[itex]\frac{-n+1}{2(2n-1)(n+1)}[/itex]|<ε.
Now take [itex]\frac{n-1}{2(2n-1)}[/itex]. Since n-1>1 when n>1 and since 4n-2 > n-1 > 1 [itex]\Rightarrow[/itex] 0<[itex]\frac{n-1}{2(2n-1)}[/itex]<1 when n>1. As long as we have n>1, we have that |[itex]\frac{-n+1}{2(2n-1)(n+1)}[/itex]|<[itex]\frac{n-1}{2(2n-1)}[/itex]. 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:
|[itex]\frac{n^{2}}{2n^{2}+n-1}[/itex]-[itex]\frac{1}{2}[/itex]|<[itex]\frac{1}{n}[/itex]<[itex]\frac{1}{N}[/itex]≤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
msell2 said:
Thanks for the help! Does this look right?

Given ε>0 we can determine an N[itex]\in[/itex]N so that |an-L|<ε for n≥N. We have:
|an-L|=|[itex]\frac{n^{2}}{2n^{2}+n-1}[/itex]-[itex]\frac{1}{2}[/itex]|=|[itex]\frac{2n^{2}}{2(2n^{2}+n-1}[/itex]-[itex]\frac{2n^{2}+n-1}{2(2n^{2}+n-1}[/itex]|=|[itex]\frac{-n+1}{2(2n^{2}+n-1}[/itex]|=|[itex]\frac{-n+1}{2(2n-1)(n+1)}[/itex]|<ε.
Now take [itex]\frac{n-1}{2(2n-1)}[/itex]. Since n-1>1 when n>1 and since 4n-2 > n-1 > 1 [itex]\Rightarrow[/itex] 0<[itex]\frac{n-1}{2(2n-1)}[/itex]<1 when n>1. As long as we have n>1, we have that |[itex]\frac{-n+1}{2(2n-1)(n+1)}[/itex]|<[itex]\frac{n-1}{2(2n-1)}[/itex].
Is that last inequality what you intended?

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:
|[itex]\frac{n^{2}}{2n^{2}+n-1}[/itex]-[itex]\frac{1}{2}[/itex]|<[itex]\frac{1}{n}[/itex]<[itex]\frac{1}{N}[/itex]≤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.
 
  • #7
What is wrong with that inequality?
 
  • #8
How are you going from
msell2 said:
[itex]\lvert \frac{-n+1}{2(2n-1)(n+1)} \rvert < \frac{n-1}{2(2n-1)}[/itex]
to
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:
[itex]\lvert \frac{n^{2}}{2n^{2}+n-1}-\frac{1}{2}\rvert < \frac{1}{n} < \frac{1}{N}≤\varepsilon[/itex]
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.
 
  • #9
n-1< 4n -2 => 1<3n which is true for n>1 get it?
 

1. What is the purpose of proving limit with epsilon and delta?

The purpose of proving limit with epsilon and delta is to provide a rigorous and precise mathematical proof for the behavior of a function at a specific point. It allows us to determine the exact value of a limit, rather than just estimating it.

2. How does the epsilon and delta definition of limit work?

The epsilon and delta definition of limit works by setting a range of acceptable values for the output of a function (epsilon) and finding a corresponding range of input values (delta) that will ensure the output stays within the epsilon range. If we can find such a delta value, then the limit of the function at that point is proven to exist.

3. What are the two main steps in proving limit with epsilon and delta?

The first step is to choose a value for epsilon and then find a corresponding value for delta that satisfies the definition of limit. The second step is to write a precise and logical proof, showing that for any value of epsilon, there exists a corresponding value of delta that satisfies the definition of limit.

4. Can you prove limit with epsilon and delta for all functions?

Yes, the epsilon and delta definition of limit can be used to prove limit for all functions, as long as the function is defined and continuous at the point in question.

5. Why is proving limit with epsilon and delta important in mathematics?

Proving limit with epsilon and delta is important because it provides a rigorous and logical framework for understanding the behavior of functions at specific points. It also helps to establish the existence and continuity of functions, which are important concepts in advanced mathematics.

Similar threads

  • Calculus and Beyond Homework Help
Replies
13
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
251
  • Calculus and Beyond Homework Help
Replies
8
Views
750
  • Calculus and Beyond Homework Help
Replies
2
Views
680
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
2K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
949
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
Back
Top