If DeltaU-->0 as DeltaX-->0 then f(x)-->L as DeltaX-->0 or as DeltaU-->0

[lamp23]

If DeltaU-->0 as DeltaX-->0 then f(x)-->L as DeltaX-->0 or as DeltaU-->0

]I am trying to prove the following theorem. I think I have the right givens and I think I just need to make the delta from the second given equal to the Epsilon from the first given so I can use the transitive property of implication.

 

[lamp23]

This is used without proof in Stewart's Calculus:

 

[daon2]

Good eye, that proof handwaves a bit. For a real proof, you will have to visit a more advanced textbook like Principles of Mathematical Analysis, W. Rudin.

You should probably view proofs in a freshman calculus textbook as "intuitive arguments", as the machinery needed for real rigor is just not present. That's not to say all are incorrect though.

As a trivial counter example to your title's question, let u(x)=0. There are other counter-examples (one in Rudin's book) but they are quite involved.

 

[lamp23]

Good eye, that proof handwaves a bit. For a real proof, you will have to visit a more advanced textbook like Principles of Mathematical Analysis, W. Rudin.

You should probably view proofs in a freshman calculus textbook as "intuitive arguments", as the machinery needed for real rigor is just not present. That's not to say all are incorrect though.

As a trivial counter example to your title's question, let u(x)=0. There are other counter-examples (one in Rudin's book) but they are quite involved.

Stewart actually does mention that this is only a proof of the special case where \(\displaystyle \Delta u \neq 0.\) He gives a proof at the end of the section.
But my question is how to prove the theorem that I hand-wrote at the top.

 

[daon2]

Stewart actually does mention that this is only a proof of the special case where \(\displaystyle \Delta u \neq 0.\) He gives a proof at the end of the section.
But my question is how to prove the theorem that I hand-wrote at the top.

The question as you wrote it is confusing, because what does the limit of f(x) as Δx -> 0 even mean? If I were to interpret it literally, the answer is just f(x) because f does not depend on the variable "Δx"..

In the "proof" given by stewart, Δy/Δx is a function of the variable Δx.

 

[lamp23]

The question as you wrote it is confusing, because what does the limit of f(x) as Δx -> 0 even mean? If I were to interpret it literally, the answer is just f(x) because f does not depend on the variable "Δx"..

In the "proof" given by stewart, Δy/Δx is a function of the variable Δx.

So I guess I should have put:
\(\displaystyle \lim_{\Delta x\to 0}f(\Delta x) = \lim_{\Delta u\to 0}f(\Delta x)\)

but I didn't put:
\(\displaystyle \lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x} = \lim_{\Delta u\to 0}\frac{\Delta y}{\Delta x} \)

because it seems like the theorem can be proved for any \(\displaystyle f(\Delta x)\)

In fact, if I'm writing a general proof I can just replace every "\(\displaystyle \Delta x\)" with a "x" and every "\(\displaystyle \Delta u\)" with a "u", right?

 

[daon2]

I'm supposing that "x->0 => u->0" gives a dependence of u on x. So now one needs to be careful... The limit as u(x)->0 may not "make sense". Consider u(x)=(x-1)^2 - 1

x->0 implies u(x)->0

But x->2 also implies u(x)->0.

So think about what the symbols "u->0" are trying to say.

It is not clear, at least to me, what "u->0" means. The limit could be considered either ambiguous or nonsensical depending on how I look at it.

 

[lamp23]

I'm supposing that "x->0 => u->0" gives a dependence of u on x. So now one needs to be careful... The limit as u(x)->0 may not "make sense". Consider u(x)=(x-1)^2 - 1

x->0 implies u(x)->0

But x->2 also implies u(x)->0.

So think about what the symbols "u->0" are trying to say.

It is not clear, at least to me, what "u->0" means. The limit could be considered either ambiguous or nonsensical depending on how I look at it.

Hmm, I'm not sure if it matters using the givens I'm using. However, Stewart defines \(\displaystyle \Delta u\) = \(\displaystyle g(x+\Delta x) - g(x)\) but I can see how my post is confusing by trying to generalize it.
I'll start a new thread using \(\displaystyle \frac{\Delta y}{\Delta x} \) instead of f(x).