# Confusion regarding delta definition of limit

1. Mar 12, 2013

### f24u7

I don't quite get the significance of the delta limit definition,

if n>N and |sn−s|< ϵ , why does the limit converges

does this simply means that there exist a number ε such that if n is great enough it will be greater than s by ε?

But this doesn't make sense, because s is the value the sequence converges to, so sncannot be greater than s

what confuses me even more is that (-1)n is a diverging sequence but
it satisfies |sn-s| < ε for ε>1

Can someone clarify this concept, thank you

2. Mar 12, 2013

### jbunniii

No, first of all you need to state the full definition: for EVERY $\epsilon > 0$ (no matter how small), there exists some $N$ such that $|s_n - s| < \epsilon$ for every $n > N$. In general, choosing a smaller $\epsilon$ means that the required $N$ will be larger.

Note that $|s_n - s| < \epsilon$ means exactly that $-\epsilon < s_n - s < \epsilon$ or equivalently $s - \epsilon < s_n < s + \epsilon$. So convergence means that no matter how small an $\epsilon$ I choose, it's possible to find an $N$ such that $s_n$ will be contained within the interval $(s-\epsilon, s+\epsilon)$ for all $n > N$.

You didn't mention what $s$ is in this case. But note that it's not enough for the inequality to be true for $\epsilon > 1$. It has to be possible to make it true for any $\epsilon > 0$, no matter how small. To see that this is impossible, take $\epsilon = 1/2$ for example, and show that there can't be any $s$ that works.