Understanding Definition of Limits

Question

I have been watching lectures on real analysis by Prof. Francis Su (Harvey Mudd College) on youtube and in the 20th lecture he defines limits of functions. He says the main idea for a function to have a limit at a point is that as the points in the domain gets closer and closer to a point, their corresponding images gets closer to a point in the co-domain. The definition of limit given by him is as follows:

$X, Y$ are metric spaces, $E \subset X$ and $p$ is a limit point of $E$.

Let $f: E \rightarrow Y.$ To say "$f(x) \rightarrow q$ as $x \rightarrow p$" or "$\lim_{x \to p} f(x) = q$" means that $\exists$ $q \in Y$ such that $\forall$ $ \epsilon \gt 0,$ $\exists$ $\delta \gt 0$ such that $\forall$ $x \in E$

$0 \lt d_X(x,p) \lt \delta \implies d_Y\left(f(x),q\right) \lt \epsilon \tag 1$

In this definition let X = Y = $\mathbb R$. I don't understand why this definition works. How does this definition reflects closeness i.e as points gets closer to $p$ their images get closer to $q$ and also why don't we say

$d_Y\left(f(x),q\right) \lt \epsilon \implies 0 \lt d_X(x,p) \lt \delta $

instead of $(1)$? If there are any mistakes in the question then please suggest.

Answer 1

I claim that $f(x)$ gets as close as you like to $q$ provided that $x$ is close enough to $p$. You challenge me to show it is within $\varepsilon$. I show that so long as $x$ is within $\delta$ of $p$ then then $f(x)$ is within $\varepsilon$ of $q$. For any $\varepsilon$ you choose, I have to be able to produce a corresponding $\delta.$

That's exactly what $(1)$ says.

Answer 2

=====second answer===

$\lim\limits_{x\to p} f(x) = q$ means

If $x$ is close to $p$ then $f(x)$ is close to $q$.

Or in other words

If $d(x,p) < \delta$ then $d(f(x), q) < \epsilon$ where $\delta$ and $\epsilon$ are small numbers.

For example. $\lim_{x\to 2}3x +5 = 11$ and we have if $|x - 2|< 1$ ($1< x < 2$) then $|f(x) - 11| < 3$ ($8< 3x+5 < 11$). And if $|x-2|< .1$ we have $|f(x)-11| < .3$ (If $1.9< x < 2.1$ then $10.7< 3x + 5 < 11.3$.)

But obviously there has to be some sort of relationship between $\delta$ and $\epsilon$.

In the examples above if $\delta = .5$ and $\epsilon = .01$ we get something that isn't true. If $x=1.75$ so $|x-2|=.25 < .5$ we get $3x + 5 = 10.25$ and $|f(x) - 11| = .75 \not < .01$.

Now this is where I think students get confused.

It seems intuitive that as we are picking $x$s to be close to $p$ and as a result the $f(x)$s are close to $q$, it would make sense that we make $\epsilon$ dependent upon $\delta$.

But it is actually the EXACT opposite.

If we choose our $\delta$ first and base our $\epsilon$ based on that, there is no need for our $\epsilon$ to actually be small.

In our examples above if we choose $\delta = 0.0000000001$ and we need $|x-2|<\delta$ to mean $|f(x) -11| < \epsilon$. Okay, let $\epsilon = 10$ million. If $|x-2| < 0.0000000001$ then $|(3x + 5) - 11|< 10$ million. Well, that is certainly true! What does that signify? Very little.

We need to show that $f(x)$ gets close to $q$. In other words, we need to show that $\epsilon$ gets small. We want to be able to show that whatever small distance we can think of, $\epsilon$ can be that distance.

In other words for any $\epsilon$ we can think of, we will be able to find a $\delta$ so that whenever $d(x,p) < \delta$ then $d(f(x),p) < \epsilon$.

And that is precisely what the definition says.

=====first answer ====

What does $\lim\limits_{x\to p} f(x) = q$ mean?

Intuition says: If we select $x$ that are really close to $p$ then the corresponding $f(x)$ will be really close to $q$.

$x$ being "really close" to $p$ means that we have "really small" distance $\delta> 0$ so that $d(xp) < \delta$. This means if we consider the values of values of $f(x)$ for all those $x$s that are within $\delta$ of $p$, we will get that all the $f(x)$s are within some value of $q$.

But that's always the case for any $q$. If $d(x,p) < \delta$ we will always have a $\underbrace{\sup d(f(x),q)}_{\text{where }d(x,p)< \delta}$ (for sake of argument this could be infinite). and $d(f(x),q)$ will be at least as small as that.

We need to get the idea that as $\delta$ gets small that $\underbrace{\sup d(f(x),q)}_{\text{where }d(x,p)< \delta}$ gets small as well.

What we want to get at is as $\delta\to 0$ then $\underbrace{\sup d(f(x),q)}_{\text{where }d(x,p)< \delta}< \delta\to 0$.

Or in other words $\lim\limits_{\delta\to 0} \underbrace{\sup d(f(x),q)}_{\text{where }d(x,p)< \delta}=0$.

... But that's circular reasoning. We are trying to define limits so we can't use limits.

We need the idea that we can "force" $f(x)$ to be close to $q$ by picking $x$s within a small distance of $p$.

We need: We can assure that $d(f(x),q) < \epsilon$ for any small distance of $\epsilon$ we want, by picking $x$s that are within some small distance $\delta$ of $p$.

Or in other words. For any $\epsilon > 0$ we can find a $\delta > 0$ so the whenever we have $d(x,p) < \delta$ we will also have $d(f(x), q) < \epsilon$.

And that's exactly what the definition is.

Answer 3

The gist of the definition is this:

Given any positive number $\epsilon$ we can get a positive number $\delta$ so that we can make the distance between $f(x)$ and $q$ become less than $\epsilon$ whenever $x$ is within a radius $\delta$ of $p.$

The condition that the given $\epsilon$ defines becomes tighter the smaller $\epsilon$ is, and the definition says no matter how small this number becomes, we can always make $f(x)$ come as close as within a radius $\epsilon$ of the number $q.$ This captures the idea of the images getting arbitrarily close to $q.$

Or, think of $\epsilon$ as a given tolerance for error; then the definition claims that no matter how accurate an approximation of $q$ is wanted, we can always get it -- that's what the $\delta$ part is all about. It simply means we can approximate $q$ as closely as we wish by approximating $p$ well enough.

This $q,$ if it exists, is then called the limiting value of the function at $x=p.$