# How does the epsilon-delta definition define a limit?

I understand what the epsilon-delta definition is saying in regards to the distance from a point c and the distance from your limit, but I don't understand how this defines a limit. Any help is appreciated. Thanks!

• It does not actually define a limit; it verifies whether L is the/a limit. You cannot use epsilon-delta to find a limit; only to test whether any L is a limit. – user99680 Mar 29 '14 at 21:32
• @user99680: How then does it verify a limit? – OpieDopee Mar 29 '14 at 21:33
• by showing that the sequence/function becomes indefinitely-close to the limit; epsilon tells you the distance to L and delta tells you how close you need to be in the domain so that the image is close to L. Then, if for every epsilon, the function can be closer than epsilon to L, L is the limit of the function. – user99680 Mar 29 '14 at 21:36
• @user99680: I almost get what you're saying, but it still escapes me. I apologize, but is there another way you can explain this? – OpieDopee Mar 29 '14 at 21:41
• let me see, please give me some time. – user99680 Mar 29 '14 at 21:42

In expressing the idea that $$\displaystyle\lim_{x\to a}f(x)=L,$$ we often use intuitive phrases such as "$f(x)$ approaches $L$ as $x$ approaches $a$" or "$f(x)$ gets closer and closer to $L$ as $x$ closer and closer to $a$". While these phrases do capture some aspects of limits, the problem with them is that they are imprecise and, in some sense, inaccurate. Hence the need for the $\epsilon$-$\delta$ definition.

Consider, for example, the statement $$\displaystyle\lim_{x\to0}\sin x=0.$$ It certainly is not always the case that $\sin x$ is getting closer to $0$ as $x$ is getting closer to $0.$ If you start with $x=\pi,$ for example, $\sin x$ equals $0$ there, but moves away from $0$ for a while as $x$ moves towards $0.$ You might say, well OK, but when $x$ is sufficiently close to $0,$ say $\lvert x\rvert<\pi/2,$ then $\sin x$ always moves towards $0$ as $x$ gets closer to $0.$ This is true, and illustrates why the precise definition of limit doesn't require that $f(x)$ always be getting closer to $L$ as $x$ gets closer to $a,$ but only that there exist some window around $a$ within which $f(x)$ is close and getting closer to $L.$ This is the role of $\delta$ in the $\epsilon$-$\delta$ definition: $\delta$ tells you the size a window around $a$ that will ensure that $f(x)$ is close and getting closer to $L$ as $x$ gets closer to $a.$

But that still isn't quite accurate, as the following example reveals. Consider the function $g(x)=x\sin(1/x).$ You've probably seen graphs of $\sin(1/x)$ and know that it oscillates back and forth between $-1$ and $1$ infinitely many times as $x$ approaches $0,$ and that the number of oscillations between $x$ and $0$ remains infinite no matter how close $x$ gets to $0.$ In fact $\sin(1/x)$ has no limit as $x$ approaches $0$ for precisely this reason. But $g(x)$ is different because of the factor $x.$ Instead of oscillating between $-1$ and $1,$ the oscillations are contained within an envelope, and furthermore, that envelope shrinks to zero width as $x$ approaches $0.$

It is intuitively reasonable in this example that $$\lim_{x\to 0}g(x)=0,$$ and we need a definition of limit that reflects that. The problem is that it is not literally true that $g(x)$ is getting closer and closer to $0$ as $x$ gets closer and closer to $0:$ even when your $\delta$-width window is very small, there are infinitely many $x$ within that window where $g(x)=0$ but then moves away from zero again. So in general we don't require that $f(x)$ always move toward $L$ as $x$ moves toward $a.$ Instead, we require that any movement of $f(x)$ away from $L$ be contained within some envelope of shrinking width. This is the role of $\epsilon$ in the $\epsilon$-$\delta$ definition: we require that eventually $f(x)$ get and stay within some window of width $\epsilon$ around $L,$ and that this be true for arbitrarily small $\epsilon.$

Putting it all together, $\lim_{x\to a}f(x)=L$ means that, for arbitrarily small $\epsilon,$ $f(x)$ stays within a distance of $\epsilon$ of $L$ provided that $x$ is sufficiently close to $a.$ By "sufficiently close", we mean within some distance $\delta$ of $a,$ where $\delta$ is allowed to depend on $\epsilon.$

Note that we don't require that $f$ be defined at $x=a,$ or, if defined, that it take the limiting value. In fact, one of the main uses for limits in introductory calculus is to understand what is happening to $f$ close to points where $f$ is undefined. So we have to exclude the point $x=a$ itself from the definition. Therefore we require that, for every $\epsilon>0,$ there exist a $\delta>0$ such that $\lvert f(x)-L\rvert<\epsilon$ whenever $\lvert x-a\rvert<\delta$ except possibly at $x=a.$

• This is very very helpful, thank you! – OpieDopee Mar 30 '14 at 13:23

To add to what user99680 said with an example-

The delta epsilon definition of a declaration of the form "if P(F, L, X) is true, then L is the limit of F at X". It is not a universal equivalence, so it leaves many cases undefined.

One of the most prominent examples is that of an infinite series. An infinite series is defined as :

$$\sum_{k=0}^{\infty} f(k) = \lim_{n \rightarrow \infty} \sum_{k=0}^{n} f(k)$$

In the case that $f$ is divergent, that leaves the limit undefined. I have a more detailed explanation in this answer, but the in short the undefined nature of the limit leads to the freedom to define things like $1 + 2 + 3 + 4 \dots = {-1 \over 12}$.