What is a plain English explanation of "$\epsilon$-$\delta$ criterion" and how to use it for proving continuity?

Question

I would prefer as little formal definition as possible and simple examples on how to proof that any given function is whether continous or not with the help of the criterion.

Answer 1

The idea of the epsilon-delta definition of a limit of a function is, in layman's terms, that you can make the value of the function as close to the limit as you like by choosing the input in a sufficiently small neighbourhood around the input value at which you're taking the limit. In particular, a simple summary of what it means for a function to be continuous is that a small perturbation in the input leads to a small perturbation in the output: if you only change $x$ by a little, then you only change $f\left(x\right)$ by a little.

A simple example of an epsilon-delta proof of the continuity of a function is proving the continuity of $$f: \mathbb{R} \rightarrow \mathbb{R},$$ $$f\left(x\right) = ax+b \qquad \left(a,b \in \mathbb{R}, \, a \neq 0\right)$$ at any point $c \in \mathbb{R}$.

We get given an arbitrary $\epsilon > 0$, and we want to find a $\delta > 0$ so that, if $|x - c| < \delta$, then $|f\left(x\right) - f\left(c\right)| < \epsilon.$ Now, $$|f\left(x\right) - f\left(c\right)| = |ax+b - \left(ac+b\right)| = |a\left(x-c\right)| = |a||x-c|,$$ so, if we pick $\delta = \frac{\epsilon}{|a|}$, then indeed $\delta > 0$ and we get $$|x - c| < \delta \iff |x-c| < \frac{\epsilon}{|a|} \iff |a||x-c| < \epsilon \iff |f\left(x\right) - f\left(c\right)| < \epsilon,$$ so by definition, $f\left(x\right)$ is continuous at $x=c$.

Answer 2

If you want a formal proof then you need to use a formal definition. It is hard to see any way out of that. It is possible that there is a definition of continuity that is equivalent to the common definition but easier to work with than the standard one but I would be surprised as it would have probably become popular.

If you want an intuitive explanation of the standard definition then I sometimes present it as a game. Suppose that I claim that $f(x)$ is continuous at $x_0$. The game is that you can challenge me with statements like this:

"Can you ensure that it will stay within small positive number of your choice of $f(x_0)$ by restricting $x$ to within small positive number of my choice of $x_0$."

If I can always succeed then I have proved that $f(x)$ is continuous at $x_0$. If I fail then it is not. Now call small positive number of your choice $\epsilon$ and small positive number of my choice $\delta$ and we are back to the usual definition.

Answer 3

A nice way of envisioning it, in my opinion, is by what my real analysis professor called "$\epsilon-\delta$ bands."

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Imagine choosing two points $x,y$ on the horizontal axis of the graph of a function $f$. Drawing vertical lines, you will find $f(x), f(y)$ where the lines cross the graph - nothing new there. We define the distance between $x$ and $y$ by $\delta$ and the distance between $f(x)$ ad $f(y)$ by $\epsilon$. Mathematically:

$$|x - y| = \delta \;\;\;\;\; \text{and} \;\;\;\;\; |f(x)-f(y)| = \epsilon$$

My professor would have said then that we have bands of these widths.

An example of what I'm talking about:

Imagine some sort of function $f$ in your mind that you know to be continuous, say $f(x) =x^2$, or the graph above (which is a slightly more complicated polynomial). Fix a point $y$ and start varying $x$. It should be apparent that as $x$ approaches $y$, both the $\epsilon$ and $\delta$ bands start to approach $0$ in size. Thus, for every $\epsilon$ band with nonzero width, we have a corresponding $\delta$ band of nonzero width.

In terms of the formal definition, then, for continuity of $f$ at $y$:

$$\forall \epsilon > 0 \; \exists \delta >0 \; \text{ such that } \; |x-y|<\delta \Rightarrow |f(x)-f(y)|<\epsilon$$

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What would it mean to be not continuous at $y$? That is the negation of the previous statement. Formally,

$$ \exists \epsilon > 0 \text{ such that } \; \forall \delta >0 \;\; |x-y|<\delta \Rightarrow |f(x)-f(y)|\geq \epsilon$$

Okay but does this mean intuitively?

Basically, we're trying to find an $\epsilon$-band of nonzero width, such that no matter what $\delta$ you use, $\epsilon$ will never get smaller than some value.

Imagine our visual example but this time $y$ is at a jump discontinuity, for example, in a function. Then - depending on how the function is defined - as your $x$ approaches $y$ from one direction, the $\delta$ band begins to approach zero width, but the $\epsilon$ band, which I remind you is $|f(x)-f(y)|$, will always be stuck at some value.

For a visual example of this, I offer the step function. It is probably one of the more basic functions you'll find with a jump discontiuity:

$$f(x)=step(x)=\left\{\begin{matrix} 0 & x<0 \\ 1 & x\geq 0 \end{matrix}\right.$$

The function I'll be showing is a vertical translation of this down by $1/2$ for clarity in the visual.

Suppose we're looking at $y = 0$ (i.e. the "jump").

Notice here how, no matter how we vary $x$, at least when $x<y$, $\epsilon$ will never be anything other than $1$. Since there does exist some $\epsilon > 0$ (namely $\epsilon = 1$, where $\epsilon$ is still $|f(x)-f(y)|$) such that no matter what $\delta$ (which is $|x-y|$) is, $|f(x)-f(y)|$ is always bigger than that $\epsilon$, the function is not continuous at $y = 0$.

Intuitively, we imagine this as sliding $x$ around, and noticing that, for $x<y$, which allows us to have a $\delta$-band of any length, the $\epsilon$-band is "locked." It's never going to change to be any value other than $1$ when $x<y$.

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3Blue1Brown on YouTube has an excellent series on "the essence of calculus." He drops a lot of the formulas and formality and focuses moreso on giving the underlying intuition of it. In Chapter 7 (https://www.youtube.com/watch?v=kfF40MiS7zA) he goes over limits and continuity. I'd recommend also checking it out.

Now, like others said, if you want to "prove" continuity, or disprove it, you still need proofs and formality. Most of the others have already touched on that, though I feel like you were seeking a more intuitive understanding.

If you want to prove continuity, you need to show that, for all $\epsilon > 0$, you have a corresponding $\delta > 0$, and this comes through by various manipulations of your function, other results like the triangle formula, and a touch of cleverness. Not much can help you there other than reading up on it and practicing it a lot.

If you want to prove discontinuity at a point, you show the negation: you show that you have a $\epsilon>0$ such that, for all $\delta>0$, you always have $|f(x)-f(y)\geq \epsilon$.