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

• I
• Hall
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.

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.

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.

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.

fresh_42
@FactChecker is right, epsilon-delta proofs are easier.

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.

PeroK
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.

Hall
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.

Hall
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##.

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

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.

Hall
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.

fresh_42 said:
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.

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.

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

PeroK
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} ##