Limit proof required for special case of chain rule

I am trying to figure out how to prove the equality I circled below in red. I have figured out how to prove the text in blue but don't know how to use that to prove the equality I circled in red.

Below I will post the givens I'm trying to use and my guess of how to prove it.

*P.S.* I understand this only proves the chain rule in the special case where $\displaystyle \Delta u \neq 0 $. This is from Stewart's Calculus and he does mention that this is not a full proof but I'm very curious how to prove this special case anyway.

http://i900.photobucket.com/albums/a...amp23/calc.jpg

I'm not sure if I'm using the right givens below.

In their blue form it looks like I will be able to use the transitive property of implication $\displaystyle (a \rightarrow b \wedge b \rightarrow c) \rightarrow (a \rightarrow c)$ if I can always let the $\displaystyle \epsilon$ from the 1st given equal the $\displaystyle \delta$ from the second line.

http://i900.photobucket.com/albums/a...3/deltau-2.jpg

Re: Limit proof required for special case of chain rule

Re: Limit proof required for special case of chain rule

Quote:

Originally Posted by

**Prove It**

I'm not looking for a way to prove the product of the limits is the limit of the product. I'm looking for a way to prove the two expressions I circled in red are equal, i.e. $\displaystyle \lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x} = \lim_{\Delta u\to 0}\frac{\Delta y}{\Delta x} $