# Thread: Exemplify why this definition of a limit does not hold

1. ## Exemplify why this definition of a limit does not hold

I've been thinking in circles over this problem so I thought I'd give this forum a try - I hope someone can help.

I've been given a definition:
For all epsilon>0 , there exists a natural number N such that if |Xn-L| < epsilon, then n is greater than or equal to N.

I need to either find an infinite sequence Xn which converges but does not satisfy the conditon above OR
find a sequence Xn and L which do satisfy the condition above but Xn does not converge to L.

I'd be grateful if anybody could talk me through what features such a sequence should have.

Thanks

2. I'll give you the sequence and it's limit from the new definition, you show that it 'converges' to that limit:

$a_n = n \ \forall n \in \mathbb{N}$ and $L = 0$
Hint: Take $N = 1$ for every $\epsilon$

After doing that, see if you can do it for any sequence.

Does this sequence guarantee $n \geq N$ when $|x_n|< \epsilon$ but ${x_n}$ does not converge?

or

Would this show that $|x_n|< \epsilon$ does not guarantee $n \geq 1$ ?

If the latter, does that apply for all possible epsilon greater than zero?

Thanks.

4. Well, is there any time where $n < N$?

5. Ah, okay we have N=1 so n will always be greater than, or equal to it.

I guess my problem is, how does this work for all possible values of epsilon? e.g what if epsilon = 1/2, i.e. for n=1,
$|1|< \frac{1}{2} => n \geq 1$

and how does this converge to 0 when n goes to infinity?

I appreciate your patience but I'm still missing something here.

6. You have (roughly speaking) a proposition of the form $P \Rightarrow Q$ where here, P means $|x_n-L| < \epsilon$ and Q means $n \ge N$
By definition, if P is false then the proposition is true, so you don't care about values of n for which $|x_n-L| \ge \epsilon$.
Now assume P is true. Is Q always true?

7. Ahh I've answered my own question Xn= n does satisfy the condition for L= 0 but does not converge to L which is what I was initially looking for.

8. Correct.