Discussion of Integral calculus

[a.s.h.]

There's no contradiction in saying h is not zero but lim _{h->0} (h) = 0. Recall also that in the definition of limits, the value of lim_{h->a}(f(h)) is independent of the value of f at a (f(a) need not even be defined for the limit to make sense).

[Boris Lagutin]

a.s.h.,

As far as I know, limit_{h->0} (h) , not simply h, is used in the definition of the derivative. By the way, when I have mentioned h, I have meant h defined by limit_{h->0}.

To sum up, my opinion is that h in the definition of the derivative is not equal to zero. So terms with h are simply ignored in the final stage of the limit calculations. Therefore, limit_{h->0} (h) is defined to be equal to zero by just convention, as I suppose. Algebraically and geometrically, h cannot be zero because a derivative expresses a slope of an original function, and a slope is a difference between two y-values divided by a difference between x-values of an original function.

P.S. Moreover, h cannot be arbitrary because it can violate the continuity of the derivative functions, as I proved it above in this thread.

Have a nice day.

[a.s.h.]

h and lim_{h->0} (h) are not the same things. The latter is the real number 0 and the former is a dummy variable (the variable that's approaching 0 for the limit). Also, it's not a convention that lim_{h->0} (h) = 0, it's something that can be proved using the epsilon-delta definition of limits.

[Boris Lagutin]

a.s.h.,

So if F(x) = x^2 is an original function, then:

F'(x) = f(x) = limit_{h->0} ((x+h)^2 - x^2)) / ((x+h) - x) = limit_{h->0} (x^2+2xh+h^2-x^2) / (x+h-x) =

the latter can be rewritten as:

= (x^2 + 2*x*(limit_{h->0} (h)) + (limit_{h->0} (h))^2 - x^2) / (x + limit_{h->0} (h) -x) =

and if limit_{h->0} (h) = 0, hence F'(x) = f(x) = (x^2 +2*x*0 + 0^2 - x^2) / (x + 0 - x) = (0/0)

If you tell that we cannot take the limit before the final stage, then, please explain why.

Thank you.

[Boris Lagutin]

a.s.h.,

There is one principal point, in my opinion, - a limit when h as a length of an interval goes to zero and a limit when h as some variable (dummy as you wrote) goes to some number (zero).
I see h in the definition of the derivative as a length shrinking to zero, not as some variable going to zero. And when I treat a slope as a limit_{h->0} (the definition of the derivative), I understand that the length of the interval between two x-values cannot be zero.

P.S. Generally speaking, all mathematics is build on convention.

Thank you.

[a.s.h.]

F'(x) = f(x) = limit_{h->0} ((x+h)^2 - x^2)) / ((x+h) - x) = limit_{h->0} (x^2+2xh+h^2-x^2) / (x+h-x) =

the latter can be rewritten as:

= (x^2 + 2*x*(limit_{h->0} (h)) + (limit_{h->0} (h))^2 - x^2) / (x + limit_{h->0} (h) -x)

This is untrue. What we would be doing here is essentially saying that lim_{h -> 0} (A(h)/B(h)) = (lim_{h->0} A(h))/(lim_{h->0} B(h)), where A(h) and B(h) are functions of h (in this case A(h) = f(x+h) - f(x) and B(h) = h, where f(t) = t^2). In other words, we are saying that the limit of a quotient is the quotient of the limits. However, this not the case if moving the limits into the numerator and denominator like this results in a 0/0. This is why we cannot do what was done above.

[a.s.h.]

I see h in the definition of the derivative as a length shrinking to zero, not as some variable going to zero. And when I treat a slope as a limit_{h->0} (the definition of the derivative), I understand that the length of the interval between two x-values cannot be zero.

P.S. Generally speaking, all mathematics is build on convention.

Thank you.

A length shrinking to 0 or a variable going to 0 is essentially the same thing. We can think of h as the displacement from the place we are taking the derivative, and limiting this towards 0. Whether we think of it as a "length" or a "variable" doesn't affect the value of the limit.

Regarding the comment of convention, I meant that "lim_{h-> 0} (h) = 0" isn't something that's true by convention, it's something that can be proved from definitions of other concepts (limits). Things like "0! = 1" or "n choose k = 0 if k > n" are conventions.

[Boris Lagutin]

a.s.h.,

Do you think that h is a variable or a function in the definition of the derivative (limit): F'(x) = (F(x+h) - F(x)) / h , as h goes to zero?

Thank you.

[a.s.h.]

a.s.h.,

Do you think that h is a variable or a function in the definition of the derivative (limit): F'(x) = (F(x+h) - F(x)) / h , as h goes to zero?

Thank you.

It's a dummy variable for a limit.

[Boris Lagutin]

a.s.h.,

Do you suppose that the limits rules for functions can be applied for h in the definition of the derivative?

Thank you.

[a.s.h.]

a.s.h.,

Do you suppose that the limits rules for functions can be applied for h in the definition of the derivative?

Thank you.

Which rules are you referring to? Remember, we can't move the limit into the numerator and denominator, because this results in 0/0, and so the limit quotient rule doesn't apply.

[Boris Lagutin]

a.s.h.,

I mean the laws of limits like the product law, sum etc. By the way, h can be only the most minimal value, in other words, h should be constant, in according to my proof of the paradox. If h is not the most minimal value, then the continuity of the derivative functions may be violated, in according to my proof of the paradox.

I must say that this discussion is great and exciting for me. I even cannot normally concentrate on my studies. I need to complete some courses yet. So I need to take "studies break"

Once again, thank you and Joe so much for such a good discussion!

Have a nice day.

[a.s.h.]

We cannot use the limit quotient rule here, since it results in a zero denominator. The quotient rule (i.e. the rule lim_{x->a} (A(x)/B(x)) = (lim_{x->a} (A(x)))/(lim_{x->a} (B(x)))) can only be used for limits where the denominator does not have 0 as limit, i.e. where lim_{x->a} (B(x)) ≠ 0.

[Boris Lagutin]

Dear a.s.h.,

Let me ask you one question. Assume F(x)=x^2 and F'(x)=f(x) (please, see the attachment). May one get a slope of the original function F(x) at some value of x if a difference between the original function two values of y is equal to zero and a difference between the original function two values of x is equal to zero?

P.S. My understanding of h in the definition of the derivative is that h goes to zero but not zero. In this way, the limit as h goes to zero is NOT undetermined on this stage: lim_{h->0} ((x^2+2xh+h^2)-x^2) / ((x+h)-x). Therefore, we can continue to calculate this limit and simply ignore terms with h at the final stage because such terms are very very small. However, terms with h cannot be ignored in infinite series (see my comments above), in my opinion. It's interesting what Isaac Newton would say about it.

Have a nice day.

[a.s.h.]

If the difference between the x-values was 0, we'd get 0/0 for the difference quotient, which is indeterminate. So we instead look at a limit as the difference between the x-values (h) tends to 0.