How does $\epsilon$-$\delta$ explain the idea of 'approach'? 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.
 A: 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.
A: 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?)
A: 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.
