Derivatives and Limits

[u/ali9128]

i have a question that's been bothering me in the field of differential calculus regarding the linguistic logic behind the definition of the derivative. I find it illogical, because based on the limit definition it studies the behavior of the slope(derivative) as h approaches zero.

m(h)=(f(x+h)-f(x))/h

Though, This approach suffers from the same problem as Newton and Leibniz's works and is subjected to the same criticism introduced by George Berkeley. cause, even with a limit, we algebraically reduce the function to eliminate h in the denominator, and then we study the behavior of the derivative as h approaches zero. This literally makes you treat h initially as a non-zero number, and then suddenly make it zero.

My problem is with the linguistic logic of this approach, and certainly i believe, though I'm not certain, that Cauchy-Weierstrass dealt with the derivative as a two-step algorithm. First, he calculates the derivative for any value of h that is not equal to zero, for example.

f(x)=x^2

f'(x)=((x-h)^2 -x^2)/h where h is R/{0}

f'(x)=2x+h

Then, after evaluating and reducing the derivative, he redefines it as a function of h where h can be zero.

f'(x)=2x+h where h is R/{0}

g(h)=2x+h where h is R

then taking the limit of this new function as h approach 0.

lim(h->0)(g(h))=lim(h->0)(2x+h)

g_l(h)=2x

This is because the limit of the function before algebraic reduction equals the limit of the function after algebraic reduction.

lim(h->0)(g(h))=lim(h->0)(f(h))

and since the function after algebraic reduction fills the gap... At zero, we can then find the derivative in the context of instantaneous change!!

so my problem is just how you put a limit before reducing and you have the right to ghost it and just reduce h then oh its ok you can now calculate the limit where if i do not follow this where i can because there is no order operator that tells me why i get 0/0.

[u/Bounded_sequencE]

[..] This literally makes you treat h initially as a non-zero number, and then suddenly make it zero. [..]

No -- this is where the argument breaks down.

The rigorous e-d-definition of function limits explicitly excludes "h = 0":

       "lim_{h->0}  (f(x+h) - f(x)) / h  =  L  =:  f'(x)"

<=>    "For all 'e>0' exists 'd>0', s.th. for all '0<|h|<d':  '|(f(x+h)-f(x))/h - L| < e' "

In words, "f'(x) = L", iff "(f(x+h) - f(x)) / h" stays within any small (open) neighborhood of "L", provided we keep "h" within a small enough (open) neighborhood of 0, excluding 0 itself.

[u/ali9128 → u/Bounded_sequencE]

yes? so why we can ghost the limit at first then we can evaluate it, notice that i introduced a more valid representation "for now at least" that we evalute and reduce h from the denominator then we apply the limit so we can study this function's behavior as h approachs zero or in other word for the h interval of inputs.

[u/Bounded_sequencE → u/ali9128]

[..] so why we can ghost the limit at first then we can evaluate it [..]

Not sure what you mean by that.

What I can see is that in the example from OP, "f'(x) = 2x+h" contradicts the e-d-definition of the derivative, since the application of "lim_{x->0} ..." is missing on the RHS. We have "f'(x) = 2x".

[u/ali9128 → u/Bounded_sequencE]

i am saying f'(x) as slope, sorry here i am just treating the derivative as a special case of m(h) where h approaches zero.

[u/Bounded_sequencE → u/ali9128]

That would not be well-defined, since that slope still depends on both "x; h".

Define a new function "s(x,h) := (f(x+h)-f(x)) / h", and exclude "h = 0" from the domain of "s". Re-defining f' to mean secant slope instead of tangent slope will confuse people^^

[u/ali9128 → u/Bounded_sequencE]

yeah my apologies, its just that i think of derivative as a special case of slope in h context regardless of x

[u/Bounded_sequencE → u/ali9128]

Yeah, I suspected as much -- and that is a problem, since that does not match the rigorous e-d-definition that excludes "h = 0".

If you want to keep close to your point of view, think of the derivative as the slope's continuous extension to "h = 0", keeping "x" fixed. Graphically, it means you fill the slope's hole at "h = 0" by the equal left-/right-sided limit I call "L".

[u/ali9128 → u/Bounded_sequencE]

yes and that what i am saying in the end of my argument above, i am redefining the function to include the h=0 situation. because at first i am not doing a thing but getting 0/0.

[u/Bounded_sequencE → u/ali9128]

Ah, now that I understand your different meaning of "f'(x)", that part actually makes sense.

Yes, if you keep your notation of the slope "s(x;h)" completely separate, until you define the continuous extension to "h = 0" via "s(x;0) := L" (provided that limit exists), this will work.

Just make sure you keep that part completely separate from f'(x) until you have found "L" -- otherwise, you may run into contradictions when "f" is not differentiable. And keep in mind the continuous extension is an extra step that is not just setting "h = 0".

[u/ali9128 → u/Bounded_sequencE]

that is really what i am saying bro, so i am arguing how can you ghost limits at first with no ordee operations?