I ##\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?

 

[fresh_42]

Do you think that I don't know that they are equivalent? It's trivial. I'm insulted.

In which case I do not understand what we are discussing. I must have misunderstood you completely and officially apologize. I only wanted to say that epsilontic isn't inevitable. It is convenient, but not the one and only possibility.

The only difference in the Weierstraß definition is that one approach is by the secants and the other by the error towards the tangent.

 

[FactChecker]

In which case I do not understand what we are discussing. I must have misunderstood you completely and officially apologize. I only wanted to say that epsilontic isn't inevitable. It is convenient, but not the one and only possibility.

The only difference in the Weierstraß definition is that one approach is by the secants and the other by the error towards the tangent.

The issue is the second part of my post. Do you think that your link is appropriate for a Freshman calculus student? (I wish that people would post their background in their profile) Without their having the background knowledge of Weierstrass, would you accept them saying that it is intuitive and calling that "rigorous"?

 

[fresh_42]

Do you think that your link is appropriate for a Freshman calculus student?

I do!

The only difference in the Weierstraß definition is that your (school math) approach is by the secants and mine (Weierstraß) by the error towards the tangent.

I do not find the epsilontic very intuitive, never did. The verbal approach I gave in post #4 is far better to understand, and by making it rigorous to find the epsilontic. Not the other way around. I like to consider the epsilontic as a consequence, not a necessity to start with.

The Weierstraß approach has essential advantages, especially for physicists. It makes it obvious why the derivative is a linear function, why it is an approximation, and what coordinates are in this context. It e.g. opens the door to linear forms. All the things that become important in physics. I cannot see what sense it makes to relearn a concept if it as well could be taught right the first place. However, that is my personal opinion.

 

[FactChecker]

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

I like that this remains within the basic properties of limits.
A beginning math student should get used to being careful with their statements:
If lim( f(x) ) and lim( g(x) ) exist and are finite, then lim( f(x)+g(x) ) = lim( f(x) ) + lim( g(x) )
If lim( f(x) ) and lim( g(x) ) exist and are finite and lim( g(x) ) $\ne$ 0, then lim( f(x)/g(x) ) = lim( f(x) ) / lim( g(x) )
I would assume that the basic properties of limits are covered early and can be referenced in a beginner's formal proof.

 

[fresh_42]

A beginning math student should get used to being careful with their statements

And we daily see where this ends up: $1/0 = \infty$ as a regular statement. I cannot see that school math was very successful with that concept. Students are kept stupid, concepts are relearned over and over again. Mathematical didactics is a huge failure.

I do not say that derivatives should be introduced via vector bundles, but epsilontic didn't succeed when I think of all these infinities in arithmetic expressions here.

 

[fresh_42]

The issue is the second part of my post. Do you think that your link is appropriate for a Freshman calculus student? (I wish that people would post their background in their profile)

This thread was and still is in the topology forum, not in a homework forum. It is labeled "I", not "B". I think that Weierstraß can be discussed here and "freshman" is not warranted.

 

[Mark44]

This thread was and still is in the topology forum

That's Topology and Analysis. You can't always trust that a thread starter picks the best forum for a question.

It is labeled "I", not "B". I think that Weierstraß can be discussed here and "freshman" is not warranted.

"I" is college level, which includes the freshman class. Taking another look at the first post in this thread, my guess is that the OP is taking freshman-level calculus.

 

[fresh_42]

That's Topology and Analysis. You can't always trust that a thread starter picks the best forum for a question.

"I" is college level, which includes the freshman class. Taking another look at the first post in this thread, my guess is that the OP is taking freshman-level calculus.

Even if. The Weierstraß definition is not over a freshman's head. I think it's even easier, but definitely not harder.

 

[FactChecker]

Even if. The Weierstraß definition is not over a freshman's head. I think it's even easier, but definitely not harder.

I am afraid that I caused a digression from the OP to derivatives.
In any case, we have presented the OP with alternatives. I think that he can see what is involved in each approach and decide what is most appropriate for him. I would caution him this way:
1) The $\epsilon, \delta$ approach stays within very basic arithmetic with very little prior development needed.
2) The Weierstrass approach is for calculus and is a good approach for more generalized (multidimensional, directional, etc.) derivative definitions.

 

[Hall]

@FactChecker is right, epsilon-delta proofs are easier.

 

[Hall]

Weierstrass decomposibility as an alternative to differentiability is, of course, not over my head, but I really cannot say anything about Freshmen. But the problem with Weierstrass decomposability is that I couldn't find source of it, nothing was available of it (given my hard work and Google fu skills). The English translation of the link given by @fresh_42 explained very little, but all I got was: a function at point $x_0$ can be decomposed as the value of the tangent line at that point plus an error function.

 

[fresh_42]

Weierstrass decomposibility as an alternative to differentiability is, of course, not over my head, but I really cannot say anything about Freshmen.

Try to view it from a physicist's point of view. A derivative in physics is a linear function in the first place, not a number. The approach via the limit of a slope yields a number, not a function. In Weierstraß's formula, the linear function is the main part of it; later specified as gradient or Jacobi matrix.

If you learn (at school) that the derivative gives a slope, and learn next that a derivative gives a gradient (first semester), then that the gradient is a linear function (second semester), and finally, that it is a section of a cotangent bundle (sixth semester), then you will have learned the same thing four times. Maybe it is a bit exaggerated to start with vector bundles, but Weierstraß's formula reduces the other three to one lesson and an easy proof that both are equivalent.

 

[FactChecker]

the problem with Weierstrass decomposability is that I couldn't find source of it,

You might not see it referenced by name, but you might see it used as the definition of the derivative in higher dimensions. As @fresh_42 says, it is the way that many (most?) people like to think of the derivative and visualize it, especially in higher dimensions. It immediately makes a lot of matrix theory and linear algebra apply straight from the definition. A person who has already spent years learning linear operators would find it painful to go back to an $\epsilon, \delta$ definition.

 

[Mark44]

The approach via the limit of a slope yields a number, not a function.

The limit of the slope at a particular point on the graph of a function yields a number, but if you do the calculation at an arbitrary point, you get a function.

Suppose that $f(x) = x^2$.

$$f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}h = \lim_{h \to 0}\frac{(x + h)^2 - x^2} h = \lim_{h \to 0} \frac{2xh + h^2}h$$
$$= \lim_{h \to 0} \frac {h(2x + h)}h = \lim_{h \to 0} \frac h h \cdot \lim_{h \to 0}2xh = 2x$$

 

[fresh_42]

The limit of the slope at a particular point on the graph of a function yields a number, but if you do the calculation at an arbitrary point, you get a function.

Suppose that $f(x) = x^2$.

$$f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}h = \lim_{h \to 0}\frac{(x + h)^2 - x^2} h = \lim_{h \to 0} \frac{2xh + h^2}h$$
$$= \lim_{h \to 0} \frac {h(2x + h)}h = \lim_{h \to 0} \frac h h \cdot \lim_{h \to 0}2xh = 2x$$

This is another reason why I do not like the limit definition. Of course, you are right and this is a function depending on the variable location. Make a survey and ask how many students are aware of the fact that the evaluation point is the variable! And how is $g(x):=x^3\, , \,g'(x)=3x^2$ a linear function?

In reality, it is far too often used when actually $\left. \dfrac{d}{dx}\right|_{x=x_0}f$ has been meant.

We can discuss this forever, and I know I'm fighting windmills. But being a minority does not change my opinion. Here is a list of ten meanings for the derivative and the word slope wasn't even among them:
https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/

This is why I am of the opinion: Teach right not twice.

 

[Mark44]

Make a survey and ask how many students are aware of the fact that the evaluation point is the variable!

I don't need to make a survey, but I have taught calculus more times than I can recall. Students who received a passing grade in the class knew that the variable was the point at which the derivative was calculated.

In reality, it is far too often used when actually $\left. \dfrac{d}{dx}\right|_{x=x_0}f$ has been meant.

Well, that's the difference between $f'(x_0)$ and $f'(x)$. Again, this is stuff I tested my students on.

 

[fresh_42]

Look at my example, $g(x)=x^3.$ Then Weierstaß's formula reads
$$
g(x_0+v)=(x_0+v)^3=\underbrace{x_0^3}_{=g(x_0)}+\underbrace{\underbrace{3x_0^2}_{=g'(x_0)}\cdot \underbrace{v}_{\text{direction}}}_{\text{slope}}+\underbrace{v^2(3x_0+v)}_{=r(v)}
$$
I simply think that this explains all relevant parts better than a limit, especially the fact that all derivatives are directional. But do not misunderstand me. I still think that the limit definition has its merits. It is just very easy to teach both and use whatever fits best in a specific situation.

 

[nuuskur]

You could prove in one fell swoop that for polynomials $f,g$ of equal degree and with leading coefficients $a,b$ respectively,
<br /> \lim _{|x|\to\infty} \frac{f(x)}{g(x)} = \frac{a}{b}<br />
using epsilon delta trickery and never have to worry about specific cases again

 

[WWGD]

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

Interestingly, that's another intuitive approach to the relation $\delta - \epsilon$, which is similar to $\lim _{ \Delta x \rightarrow 0} \frac { \Delta f}{ \Delta x}$