How does $\epsilon$-$\delta$ explain the idea of 'approach'?

Question

If $\lim_{x\to x_0}f(x)=L$

I Informally the idea of limit is that as $x$ approaches $x_0$, the value that $f(x)$ approaches is its limit at $x_0$.

II The $\epsilon$-$\delta$ definition states that- $\forall \epsilon$, $\exists \delta$, such that:
$0<|x-x_0|<\delta\Rightarrow|f(x)-L|<\epsilon$. For any value of $\epsilon$ if I can provide a $\delta$, then $L$ is the limit of $f(x)$ at $x_0$.

In other words for any distance from $L$, that has been provided, I have to find a distance, from $x_0$, such that for all the values of $x$ closer to $x_0$, their respective $f(x)$s are closer to $L$.

In many of the calculus related videos that I watch, the informal definition (The approach definition) is used to prove theorems, and other things. Right now I'm only dealing with single variable calculus, so I can imagine for a 2D graph how values of $f(x)$ can approach $L$, as $x$ approaches $x_0$. But what about in a multi-variable setting.

So how does the $\epsilon$-$\delta$ definition explain the 'approach' idea, and how is the 'approach' idea of limit always true?
I've also read a statement that $\epsilon$-$\delta$ does not actually define the limit, but is instead just a way to prove a limit is true or false?

This is one statement in a proof I saw: If $\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}=f'(x)$, assuming $\frac{f(x+h)-f(x)}{h} = f'(x) + \sigma(h)$, where $\sigma(h)$ is a 'junk term' that approaches $0$ as $h$ approaches $0$. The proof was based on this assumption.
Mathematically when I apply limit on both sides I get $\lim_{h \to 0} \sigma(h)=0$. But I'm still skeptical if this (assuming function equal to limit + 'junk') can be done for any case.

Answer 1

Well, nothing is actually "moving".

But as you take values of $x$es that are "very near" $a$ then resulting values of $f(x)$es will in response be "very near" $L$. And if you restrict or range of $x$es by much narrower and restrict you $x$es to be closer to $a$ that will result in the range of $f(x)$ values to be closer to $L$.

So $\lim_{x\to a} f(x) = L$ means for whatever target range we desire to hone in toward $f(x)\to L$ we'd like... that is to say if we want all of our $f(x)$ outputs to all be within $L' < f(x) < L''$ for values of $L'$ and $L''$ as close to $L$ as we practically like. We can find cooresponding $a', a''$ so that whenever we choose $a' < x < a''$ then the result will be that $f(x)$ will be $L' < f(x) < L''$.

So how do we express that analytically?

Our target is that we want $L', L''$ to arbitrarily as close as a value of $\epsilon$ within $L$. That is for example, if we want $L',L''$ to be within one-hundred billionth of a unit of $L$ that is $\epsilon = $ one-hundredth billion of a unit and our goal is that the $x$ we pick (that will be somewhere near but not equal to $a$) will result in $L - \epsilon = L -\frac 1{100000000000} < f(x) < L + \frac 1{100000000000} = L + \epsilon$.

Or in other word $|f(x) - L| < \epsilon$.

That's our target goal. And we want to achieve that be picking $x$ close to $a$.

That is to say.... If $\lim_{x\to a} f(x) = L$ that says we can, by restricting our choices of $x$ to be very very very close to $a$ have the result that $|f(x)-L| < \epsilon$ and no matter how small we choose $\epsilon$ we can always reach this goal be pick $x$ closer and closer to $a$.

How close?

Well for every target distance of $\epsilon$ that we want $f(x)$ to be within $L$. there is $\delta$ that will be a sufficient requirement so that if $a' = a-\delta$ and $a'' = a+ \delta$ where $\delta$ that by choosing $x$ so that $a' = a-\delta < x <a+\delta < a''$ that guarantees that $L-\epsilon < f(x) < L + \epsilon$.

Now what if we can't guarantee that result? Well... then we failed in our attempt of assuring we can hone in on $f(x)\to L$ by choosing $x\to a$. If we fail we fail and we can't do it. But to say that can force $f(x)$ to be close to $L$ by forcing $x$ to be close to $a$ means precisely:

For any target goal of getting $f(x)$ within a small $\epsilon$ range of $L$, we can find a small $\delta$ range (specifically with that $\epsilon$ in mind) so that choosing any $x$ within that $\delta$ near $a$, assures that the responding $f(x)$ will be within $\epsilon$ of $L$.

That is what a $\delta$-$\epsilon$ construction means.

......

Maybe a practical example:

Suppose want to show that $\lim_{x\to 2} x^2 = 4$.

So we want to show we can get $x^2$ close to $4$ but picking $x$ close to $2$.

How close?

Well we can get $x^2$ within $5$ units of $4$, That is $-1=4-5 < x^2 < 4+5 = 9$ by choosing $x$ within $1$ unit of $2$. That is if $1=2-1 < x < 2+1 = 3$ then $1^2 < x^2 < 9$ and we are within the target range.

Well, that's not very close. Let's try to get closer. Let try to get $x^2$ withing $1$ unit. That is we want $3= 4- 1 < x^2 < 4+ 1$. And we can do that be taking $x$ closer to $2$.

How close? Well if $x$ is within $\frac 14$ unit of $2$ then $1\frac 45 < x < 2\frac 15$ and $3 < 3.24=(1\frac 45)^2 < x^2 <(2\frac 15)^2 = 4.84 < 5$ and we are within range.

Still can we get closer?

Let's see if we can get $x^2$ within an $\epsilon = \frac 1{100,000,000,000}$ of $4$ That is can we get $3.99999999999 < x^2 < 4.000000000001$? by get $x$ close to $2$?

How close.

Well, if we choose $\delta = \frac 2{1,000,000,000,000}$ then $2- \delta = 1.999999999998 < x < 2.000000000002=2+\delta$ will result in $3.999999999992000000000004 < x^2 < 4.000000000008000000000004$ and that is within our desired range.

So, so far, for every target goal of getting $x^2$ within some small $\epsilon$ of $4$ we've succeeding but taking $x$ within some some $\delta$ of $2$.

Can we do that for any $\epsilon$?

For any $\epsilon$ will we always be able to find a $\delta$ so that taking $x$ within $\delta$ of $2$ will guarantee that $x^2$ will be within $\epsilon$ of $4$?

Well.... gosh.... I sure hope so...... I mean.... I'm putting my reputation out there by claiming as $x\to 2$ we have $x^2 \to 4$. I'd better be able to back that up.... so.... gosh, I'd better be able to claim that.... (guess I'm not going to sleep very well tonight if I can't....)

Answer 2

The way I look at it is this: there are two sort of Mathematical arguments you can make: informal arguments, where you don't care about sticking very close to definitions and instead make intuitive arguments - and formal arguments, where we care a lot about making sure we stick to definitions very carefully.

The "approach" idea of a limit you speak of is informal. It's not a definition. It's simply trying to show you what the essence of a limit is.

The ${\epsilon-\delta}$ definition is just that - a definition. This is rigorous. This is what you need to use in formal Mathematics to prove limits.

Trying to prove the two above ideas are the same doesn't really make sense. ${\epsilon-\delta}$ is a definition trying to formally define what is meant by a limit, and that "approach" idea is just giving you a taste of what the definition says.

Answer 3

Short not quite an answer. We don't explain "approaches", we do without it, by carefully defining something equivalent we can reason with.

When you think of limits in terms of "approaching" you run into philosophical questions about time passing or numbers being infinitely close together.

Mathematicians have decided to replace vague notions of "infinitely close" by requiring "infinitely many" inequalities. That's what you do when you argue that "for every $\epsilon$ ...".

Added after the question was edited:

It's routine algebra to show that these two statements say the same thing about a function $f$ and a number $L$: $$ \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = L $$ and $$ \lim_{h \to 0} \left( \frac{f(x+h) - f(x)}{h} - L \right) = 0 . $$ If you write $\sigma(h)$ for the expression in the large parentheses in the second equation then showing that $L$ is the derivative of $f$ at $x$ is just the same as showing $\sigma$ has limit $0$ as $h \to 0$. It's an algebraic trick that's sometimes useful.

Answer 4

The word "approach" does not serve well in a mathematical definition. I would argue that it is misleading to say that if $\lim_{x\to x_0} f(x) = L$ then $f(x)$ approaches $L$ as $x$ approaches $x_0.$

Let's define a function $f$ by $$ f(x) = \begin{cases} x \sin \frac1x & x \neq 0, \\ 0 & x = 0. \end{cases} $$

You should be able to confirm by a $\delta$-$\epsilon$ proof that $$ \lim_{x\to 0} f(x) = 0. $$ For any $\epsilon,$ one choice of a suitable $\delta$ is to set $\delta = \epsilon.$ Just remember that $-1 \leq \sin\frac1x \leq 1$ and therefore $\lvert x\sin\frac1x \rvert \leq \lvert x \rvert.$

But is is really OK to say that $f(x)$ approaches $0$ as $x$ approaches $0$? The way I see it, if you decrease $x$ continously down toward $0,$ the value of $f(x)$ is sometimes approaching $0,$ but it infinitely often actually reaches $0$ and then starts going away from $0.$ It just keeps approaching, going away, approaching, going away, back and forth infinitely many times. It's not even possible to say whether $f(x)$ finally approaches $0$ from above or below, because there is no final approach. There are just infinitely many oscillations of the function that inexorably get squeezed down to the limit $0$.

So I would argue that the purpose of $\delta$-$\epsilon$ is not to explain the "approach" idea, because the "approach" idea of the limit is not always true.

(That is, it is not always true unless you have a very technical definition of the word "approach" that means "satisfies the $\delta$-$\epsilon$ definition"; but that seems silly to me. We have a definition of "limit" expressed by $\delta$-$\epsilon$ without using the word "approach", so why introduce another word that we have to interpret with a special meaning that contradicts its plain English meaning?)