# How does existence of epsilon - delta relation proves that limit exists?

This is how I understand epsilon ($\epsilon$) and delta ($\delta$) relation (please correct me if I'm wrong)-

Let the limit of $f(z) = L$ as $z$ approaches $k$. To prove or see that it is actually the limit,

let us take an $\epsilon$ such that $\epsilon > |f(z) - L |$

If, for every $\epsilon>0$, there is a $\delta>0$ such that $\delta> |z-k|$ , then the limit exists and is equal to $L$.

I'm unable to understand how this works. How does the existence of such a $\delta$ proves that limit is what we assumed?

• Generally the way you prove such statements is first you declare an $\epsilon>0$. Then you try to find a $\delta>0$ (this is typically the tricky part) such that if $|z-k| < \delta$ then you can show that $|f(z) - L|<\epsilon$. You don't state first that $|f(z) - L| < \epsilon$ and then try to find a $\delta$. – Mnifldz Mar 24 '15 at 21:47
• In words the definition says "$f(z)$ can be made arbitrarily close to $L$ by making $z$ sufficiently close to $k$". Does that help? – Ian Mar 24 '15 at 22:05
• Thank you for the edit and I think I have got what the definition mean. I'm confused about why is it so. – Chand Sethi Mar 24 '15 at 23:05

I once read something like this in a Serge Lang book with the sequence $\frac{1}{n}.$ If the terms cannot be made as small as desirable by choosing large enough $n,$ then there is some $\epsilon > 0$ such that, no matter how large $n$ may be, we have $\frac{1}{n} \geq \epsilon.$ Therefore, we have $n \leq \frac{1}{\epsilon}.$ Then if we let $N= \lfloor \frac{1}{\epsilon} \rfloor + 1,$ then $n \leq N$ for all integers, and so $N$ is the largest integer, which is absurd. I found that thinking about this argument helps immensely in understanding why the definition is the way it is.
With regards to the game analogy, what is really revealing is to think about what happens when Player 1 wins. Suppose he did choose an $\epsilon$ so that, no matter how small the $\delta$ Player 2 chooses, he cannot get $f(z-\delta,z+\delta)$ to lie inside $(L-\epsilon,L+\epsilon).$ Thus, there is a point 'infinitely close' to $z$ that gets separated from $z$ by a distance of at least $\epsilon.$ It seems that $f$ creates a 'tear' at $z.$
Your problem is that you need to realize that the existence of a suitable $\delta$ corresponding to any arbitrary $\epsilon>0$ such that whenever $|z-k| < \delta_\epsilon$ you have $|f(z)-L| < \epsilon$ is the definition of the statement that the limit of $f(z)$ as $z \to k$ is $L$.
For any given $f$ and $L$ it may be hard to find a suitable $\delta$ for each $\epsilon$ and may be hard to prove that this formula for $\delta$ always works to make the function close to $L$. But once you have shown this, you need go no further because you have shown that the definition of "limit of $f(z)$" is satisfied by $L$.
It's like a game: Without saying what $\epsilon$ is (other than that it is positive), make an educated guess for a good $\delta(\epsilon)$, prove that when $z$ is that close to $k$ the function is within $\epsilon$ of $L$ and you win. (Assuming you consider yourself "winning" by proving that $L$ is the limit.)