##\epsilon - \delta## proof and algebraic proof of limits

[Hall]

It occurred to me that I should ask this to people who passed the stage in which I’m right now, being unable to find anyone in my milieu (maybe because people around me have expertise in other fields than mathematics) I reckoned to come here.

Let’s see this sequence: $s_n = \frac{3n+1}{7n-4}$. We need, not need I mean simply we... umm... we “would like to” find its limit.

I do this when I’m presented with dishes like that:
$s_n = \frac{3n+1}{7n-4}$, for very large $n$ that “1” and “4” won’t matter much, and that’s something very natural. So, for large $n$, we actually have
$$
s_n = \frac{3n}{7n}= \frac{3}{7}$$
Thus, $\lim ~ s_n = \frac{3}{7}$.

$\epsilon-\delta$ proof:
Consider a sufficiently small $\epsilon \gt 0$.

$\big| \frac{3n+1}{7n-4} -\frac{3}{7} \big| \lt \epsilon$

$\big| \frac{19}{7(7n-4)} \big| \lt \epsilon$

As $n \in \mathbf{N}$, we have $7n-4 \gt 0$. Therefore,

$\frac{19}{7(7n-4)} \lt \epsilon$

$\frac{19}{49 \epsilon} + \frac{4}{7} \lt n$

Take $N = \frac{19}{49 \epsilon} + \frac{4}{7}$. By reversing the steps, we can establish the famous statement

$n\gt N \implies \big| \frac{3n+1}{7n-4} - \frac{3}{7} \big| \lt \epsilon$

For good reasons, the first workout is more rigorous to me, I’m sure you will object “what did you really mean that 1 and 4 won’t matter much for large n?”, but that’s something we can perceive , I’m not advocating for empiricism but that $\epsilon-\delta$ proof doesn’t really tell me anything, it simply says “you can make $s_n$ as close to $3/7$ as you like”.

My argument that for very large n 1 and 4 won’t matter much, is something like the limitation of my eyes; if a helicopter were to take me slowly upwards: I would, surely, say shrubs don’t matter much and after a sufficient height I would really be unable to see those little shrubs and so, the only things that would be visible to me are trees.

If you could provide me the actual philosophy of Augustine Cauchy behind his “as close as you like”, the philosophy that he might have conveyed in those polytechnic college lectures, I would be grateful.

 

[PeroK]

It's not always as obvious as this. What do you do when come to harder limits? Or, to prove a theorem about limits where there is no specific sequence to look at?

 

[FactChecker]

Your point is understandable, but the $\epsilon, \delta$ proof is the gold standard. Your thinking is ok in cases where there is nothing treacherous, but that is not guaranteed. A common example is the ratio of two values that both approach zero. Then the question is which one approaches zero faster. In that case, your intuitive approach gets complicated fast. (The $\epsilon, \delta$ proofs can also get complicated, but they remain rigorous, whereas an intuitive "proof" gets vague and debatable.)

 

[fresh_42]

It occurred to me that I should ask this to people who passed the stage in which I’m right now, being unable to find anyone in my milieu (maybe because people around me have expertise in other fields than mathematics) I reckoned to come here.

Let’s see this sequence: $s_n = \frac{3n+1}{7n-4}$. We need, not need I mean simply we... umm... we “would like to” find its limit.

I do this when I’m presented with dishes like that:
$s_n = \frac{3n+1}{7n-4}$, for very large $n$ that “1” and “4” won’t matter much, and that’s something very natural. So, for large $n$, we actually have
$$
s_n = \frac{3n}{7n}= \frac{3}{7}$$
Thus, $\lim ~ s_n = \frac{3}{7}$.

$\epsilon-\delta$ proof:
Consider a sufficiently small $\epsilon \gt 0$.

$\big| \frac{3n+1}{7n-4} -\frac{3}{7} \big| \lt \epsilon$

$\big| \frac{19}{7(7n-4)} \big| \lt \epsilon$

As $n \in \mathbf{N}$, we have $7n-4 \gt 0$. Therefore,

$\frac{19}{7(7n-4)} \lt \epsilon$

$\frac{19}{49 \epsilon} + \frac{4}{7} \lt n$

Take $N = \frac{19}{49 \epsilon} + \frac{4}{7}$. By reversing the steps, we can establish the famous statement

$n\gt N \implies \big| \frac{3n+1}{7n-4} - \frac{3}{7} \big| \lt \epsilon$

For good reasons, the first workout is more rigorous to me, I’m sure you will object “what did you really mean that 1 and 4 won’t matter much for large n?”, but that’s something we can perceive , I’m not advocating for empiricism but that $\epsilon-\delta$ proof doesn’t really tell me anything, it simply says “you can make $s_n$ as close to $3/7$ as you like”.

You are wrong. Your first attempt was indeed more rigorous than the second. The second has a serious flaw: you cannot say "by reversing the steps", you have to do it! It is not clear whether it can be reversed at all since we used inequalities. Assume an $\varepsilon >0,$ define a number $N=N(\varepsilon )$ and prove that
$$
\left|\dfrac{3n+1}{7n-4}-\dfrac{3}{7}\right| < \varepsilon \text{ for all }n>N
$$
Leaving it to the reader is not the right thing to do. Your proof by words was much more rigorous.

My argument that for very large n 1 and 4 won’t matter much, is something like the limitation of my eyes; if a helicopter were to take me slowly upwards: I would, surely, say shrubs don’t matter much and after a sufficient height I would really be unable to see those little shrubs and so, the only things that would be visible to me are trees.

If you could provide me the actual philosophy of Augustine Cauchy behind his “as close as you like”, the philosophy that he might have conveyed in those polytechnic college lectures, I would be grateful.

Put it another way. If a sequence converges to a limit, say $L$, then all but finitely many sequence members are in a radius of, say $1/2$ around $L$. The same is true for $1/3$ or whatever number $\varepsilon >0$ I choose. There are always infinitely many sequence members closer than $\varepsilon$ to $L$ and only finitely many outside of this neighborhood of radius $\varepsilon$ around $L$. The $\varepsilon - N$ definition only formalizes that.

 

[pbuk]

I think there is a language barrier here. There is no rigour in this "proof":

$s_n = \frac{3n+1}{7n-4}$, for very large $n$ that “1” and “4” won’t matter much, and that’s something very natural. So, for large $n$, we actually have
$$
s_n = \frac{3n}{7n}= \frac{3}{7}$$
Thus, $\lim ~ s_n = \frac{3}{7}$.

, the reason that you find this more appealing is because you find it more intuitive: this is a subjective judgement.

Rigour is objective: a proof cannot be "more rigorous to me". If you correct the flaw in your $\varepsilon, \delta$ proof pointed out by @fresh_42, then it will be a rigorous proof whereas your first argument will never be rigorous.

 

[fresh_42]

... whereas your first argument will never be rigorous.

I don't agree here. His first argument is rigorous in my mind. Written in formulas it goes
\begin{align*}
\dfrac{3n+1}{7n-4}=\dfrac{3n+O(1)}{7n+O(1)} =\dfrac{3}{7}+O(n^{-1}) \longrightarrow \dfrac{3}{7}
\end{align*}

 

[PeroK]

It can be made rigorous (based on known results):
$$\lim_{n \to \infty} \frac 1 n = \lim_{n \to \infty} \frac 4 n = 0$$$$\Rightarrow \lim_{n \to \infty} (3 +\frac 1 n) = 3, \ \lim_{n \to \infty} (7 - \frac 4 n) = 7$$$$\Rightarrow \lim_{n \to \infty} \frac{3 +\frac 1 n}{7 - \frac 4 n} = \frac 3 7$$$$\Rightarrow \lim_{n \to \infty} \frac{3n + 1}{7n - 4} = \frac 3 7$$

 

[pbuk]

I don't agree here. His first argument is rigorous in my mind. Written in formulas it goes
\begin{align*}
\dfrac{3n+1}{7n-4}=\dfrac{3n+O(1)}{7n+O(1)} =\dfrac{3}{7}+O(n^{-1}) \longrightarrow \dfrac{3}{7}
\end{align*}

I agree that is rigorous, however you have done more than translate from words to symbols, you have for instance changed the OP's incorrect $... = \dfrac{3}{7}$ to $... \longrightarrow \dfrac{3}{7}$. In my mind this makes your argument a different one.

 

[fresh_42]

I mainly wanted to emphasize that a proof does not have to follow the epsilontic to be rigorous. Mathematicians wrote their proofs in a verbal form for centuries, before we all got used to Bourbaki's efficiency. It is a priori not wrong to perform a proof with words. It is only so much more difficult since language is ambiguous and there are often so many cases involved.

 

[pbuk]

I mainly wanted to emphasize that a proof does not have to follow the epsilontic to be rigorous.

Agreed, however that does not help the OP's confusion between rigour and intuitiveness.

 

[FactChecker]

I mainly wanted to emphasize that a proof does not have to follow the epsilontic to be rigorous. Mathematicians wrote their proofs in a verbal form for centuries, before we all got used to Bourbaki's efficiency. It is a priori not wrong to perform a proof with words. It is only so much more difficult since language is ambiguous and there are often so many cases involved.

There are a great many things about arithmetic with orders that need to be established before this can be called "rigorous". I am willing to bet that the OP does not come anywhere close to that. In fact, the definition of "limit" is probably in $\epsilon, \delta$ form and any connection to orders is purely wishful thinking at this point in his development.

 

[FactChecker]

What is he going to do about $\frac {f(x)-f(x_0)}{x-x_0}$ with this approach?

 

[fresh_42]

What is he going to do about $\frac {f(x)-f(x_0)}{x-x_0}$ with this approach?

What Weierstraß did:
\begin{equation} \mathbf{f(x_{0}+v)=f(x_{0})+J(v)+r(v)} \, , \,\mathbf{r(v)}=\omega(\mathbf{v})\end{equation}

 

[FactChecker]

What Weierstraß did:
\begin{equation} \mathbf{f(x_{0}+v)=f(x_{0})+J(v)+r(v)} \, , \,\mathbf{r(v)}=\omega(\mathbf{v})\end{equation}

IMHO, this is several levels of sophistication and abstraction above the typical beginner with limits.
But I can not look up any information about his mathematical background (as usual).

 

[fresh_42]

IMHO, this is several levels of sophistication and abstraction above the typical beginner with limits.
But I can not look up any information about his mathematical background (as usual).

Maybe all of that, but it shows the intuition behind the concept. Both of my examples come down to a remainder term that converges to zero. And getting smaller and smaller isn't hard to understand. One does not need the epsilontic, although it makes things easier. I described in post #4 what the epsilontic is actually good for, but you do not need it. As I said, pre-Bourbaki isn't so long ago.

This discussion feels a bit as if the majority here would reject the Principia as well as the Disquisitiones for lack of rigor.

 

[PeroK]

It occurred to me that I should ask this to people who passed the stage in which I’m right now, being unable to find anyone in my milieu (maybe because people around me have expertise in other fields than mathematics) I reckoned to come here.

The answer is not to worry about $\epsilon-\delta$; and, if your analysis professor tries to fail you, refer him or her to @fresh42 (or summon the ghost of Karl Weierstrass).

 

[FactChecker]

The answer is not to worry about $\epsilon-\delta$; and, if your analysis professor tries to fail you, refer him or her to @fresh42 (or summon the ghost of Karl Weierstrass).

Or he can actually learn $\epsilon-\delta$ for his classes and use other methods for his own work. It is not that hard and most classes quickly leave that behind.

 

[PeroK]

Or he can actually learn $\epsilon-\delta$ for his classes and use other methods for his own work. It is not that hard and most classes quickly leave that behind.

I like the $\epsilon-\delta$ definition, but I feel proofs from first principles are overused and instead more use should be made of theorems and previous results for limits. Take, for example, the proof that:
$$\lim_{x \to a} f(x) = F \ \text{and} \ \lim_{x \to a} g(x) = G \ \Rightarrow \ \lim_{x \to a} f(x)g(x) = FG$$There is a typical proof here:

https://math.stackexchange.com/ques...s-equals-the-product-of-the-limits-is-this-pr

This proof seems to me rather tangled.

The alternative is to break this down into a series of bite-sized results:

First, it's easy to show (using $\epsilon-\delta$) that:
$$\lim_{x \to a} f(x) = 0 \ \text{and} \ \lim_{x \to a} g(x) = 0 \ \Rightarrow \ \lim_{x \to a} f(x)g(x) = 0$$Next, to return to the main proof, it follows that:
$$\lim_{x \to a} (f(x) - F) = 0 \ \text{and} \ \lim_{x \to a} (g(x) - G) = 0$$Hence:
$$\lim_{x \to a} (f(x) -F)(g(x)-G) = 0$$And:
$$\lim_{x \to a} (f(x)g(x) -Fg(x)-Gf(x) + FG) = 0$$Finally, we use the basic property of limits:
$$\lim_{x \to a} Fg(x) = F\lim_{x \to a} g(x) = FG, \ \text{and} \ \lim_{x \to a} Gf(x) = FG$$To get the result.

IMO, this is a more minimal use of $\epsilon-\delta$ to get some basic results, which can then be combined and thereby avoid the complexities of the above proof.

 

[Hall]

"When the successive numerical values of such a variable decrease indefinitely, in such a way as to fall below any given number, this variable becomes what we call infinitesimal, or an infinitely small quantity."

That man really had something. What an expression "to fall below any given number"!

 

[Mark44]

After the basic concept of the limit is defined in many textbooks (via $\delta$ and $\epsilon$), several limit properties are developed, including limit of a sum, limit of a product, and limit of a quotient. Problems such as the one shown in this thread can be solved by the application of these limit properties.
$\lim_{n \to \infty}\dfrac{3n + 1}{7n - 4} = \lim_{n \to \infty} \dfrac n n \cdot \dfrac{3 + 1/n}{7 - 4/n}$
$= \lim_{n \to \infty}\dfrac n n \cdot \lim_{n \to \infty} \dfrac{3 + 1/n}{7 - 4/n} = 1 \cdot \dfrac 3 7 = \dfrac 3 7$.

From the 2nd step on, each of the limits shown can be proved with a $\delta$ and $\epsilon$ argument, if necessary.

 

[FactChecker]

Maybe all of that, but it shows the intuition behind the concept. Both of my examples come down to a remainder term that converges to zero. And getting smaller and smaller isn't hard to understand. One does not need the epsilontic, although it makes things easier. I described in post #4 what the epsilontic is actually good for, but you do not need it. As I said, pre-Bourbaki isn't so long ago.

This discussion feels a bit as if the majority here would reject the Principia as well as the Disquisitiones for lack of rigor.

It may just be the bias of my background. When I see ${lim}_{x->x_0} \frac{f(x)-f(x_0)}{x-x_0}$, I feel comfortable with issues of how to determine existence , uniqueness, etc. I feel that I know how to determine if a limit exists and is unique directly from the $\epsilon, \delta$ definition.
I don't feel so comfortable with issues of existence and uniqueness in the Weierstrass representation. Those issues might even require me to resort to $\epsilon, \delta$ definitions.

 

[fresh_42]

I like the intuition behind Weierstrass's formula:

If I change the function at $x_0$ a bit in direction of $v$, i.e. $f(x_0+v)$, then it is close, up to an error $r(v)$ that converges faster to zero than $v$, to where I started from $f(x_0)$ plus a step on the tangent $J_{x_0}(v)$ at $x_0$ in direction $v$.

The formula expresses exactly what happens geometrically.

 

[FactChecker]

I like the intuition behind Weierstrass's formula:

If I change the function at $x_0$ a bit in direction of $v$, i.e. $f(x_0+v)$, then it is close, up to an error $r(v)$ that converges faster to zero than $v$, to where I started from $f(x_0)$ plus a step on the tangent $J_{x_0}(v)$ at $x_0$ in direction $v$.

The formula expresses exactly what happens geometrically.

That is intuitive, but things like "converges faster" and "tangent $J_{x_0}(v)$" are yet to be defined. Their definition might even require an $\epsilon, \delta$. So I would say that it is not yet a rigorous definition.

 

[fresh_42]

That is intuitive, but things like "converges faster" and "tangent $J_{x_0}(v)$" are yet to be defined. Their definition might even require an $\epsilon, \delta$. So I would say that it is not yet a rigorous definition.

They are fully equivalent.

 

[FactChecker]

They are fully equivalent.

Those phrases are very vague and not defined in that statement. They are NOT definitions. They are intuitive, nothing else. They can be made rigorous, but that becomes more complicated than your simple statement.

 

[fresh_42]

Those phrases are very vague and not defined in that statement. They are NOT definitions. They are intuitive, nothing else. They can be made rigorous, but that becomes more complicated than your simple statement.

What do you mean? That the definitions via the limit of the slope and Weierstrass's decomposition formula are equivalent, or my presentation of the latter. The first is rigorous (proof on Wikipedia), and the second is correct. I did not define it properly, we were simply talking about alternatives. There wasn't a need to retype the Wikipedia page.

 

[fresh_42]

I have found a nice historical essay about the matter:

The Changing Concept of Change: The Derivative from Fermat to Weierstrass
by Judith v. Grabiner (UCLA)

https://www.maa.org/sites/default/files/0025570x04690.di021131.02p02223.pdf

 

[FactChecker]

I doubt that a complete definition of the derivative can be made using the Weierstrass definition without a lot more discussion than the essentially self-contained $\epsilon, \delta$ definition. You would have to define the tangent in a way that existence and uniqueness can be determined. Also, you would have to define "converges faster" in terms of order.
But going back to the original question of the definition of "limit", I could have (should have?) used a more general example of $\frac{f(x)-f(x_0)}{g(x)-g(x_0)}$, or other examples, where the $\epsilon, \delta$ approach is much more basic than trying to compare the orders of (in this case) the numerators and denominators.

 

[fresh_42]

I doubt that a complete definition of the derivative can be made using the Weierstrass definition without a lot more discussion than the essentially self-contained $\epsilon, \delta$ definition. You would have to define the tangent in a way that existence and uniqueness can be determined. Also, you would have to define "converges faster" in terms of order.
But going back to the original question of the definition of "limit", I could have (should have?) used a more general example of $\frac{f(x)-f(x_0)}{g(x)-g(x_0)}$, or other examples, where the $\epsilon, \delta$ approach is much more basic than trying to compare the orders of (in this case) the numerators and denominators.

Your doubt is fortunately irrelevant. The $\varepsilon$ part remains in $\lim_{v \to 0}\dfrac{r(v)}{\|v\|}=0.$

Here is the proof:
https://de.wikipedia.org/wiki/Weierstraßsche_Zerlegungsformel

Google Chrome can translate it for you. I am too lazy to re-type the obvious.

 

[FactChecker]

Your doubt is fortunately irrelevant. The $\varepsilon$ part remains in $\lim_{v \to 0}\dfrac{r(v)}{\|v\|}=0.$

Here is the proof:
https://de.wikipedia.org/wiki/Weierstraßsche_Zerlegungsformel

Google Chrome can translate it for you. I am too lazy to re-type the obvious.

Do you think that I don't know that they are equivalent? It's trivial. I'm insulted. I do not know the background of the OP since he didn't put it in his profile. If he is a beginning calc student, do you think that your link is appropriate?