Hi!
could someone explain how Δx heads toward 0?
As I don't see how we get from 2x+Δx to simply 2x
I was trying to follow the examples on the derivatives page here:
http://www.mathsisfun.com/calculus/derivatives-introduction.html
Big thanks! Tim.
"Heads toward 0" is descriptive, but not very technical! We would rather say,
"In the limit as Δx approaches 0," written \(\displaystyle \displaystyle \lim_{\Delta x \to 0}(2x + \Delta x)\)
To define that limit, let me ask "How close do you have to get?" Suppose there is some incredibly small number, usually called \(\displaystyle \epsilon\), such that if you are
that close to \(\displaystyle 2x\) then the difference doesn't matter any more. I will then say that
no matter how small you choose \(\displaystyle \epsilon\), I can find a value of \(\displaystyle \Delta x\) that will be
even closer to \(\displaystyle 2x\). In the specific limit you are looking at, I just have to make \(\displaystyle \Delta x < \epsilon\). Thus without ever actually setting \(\displaystyle \Delta x =0\), I can meet your challenge to get arbitrarily close to \(\displaystyle 2x\).
This game of challenge and response is sometimes called "epsilonics." Have fun with it!
