Then you are making a very fundamental mistake in taking limits. Taking limits you do NOT "substitute 0 into \Delta x". That's the whole point of limits. A limit is NOT just a fancy way to talk about evaluating a function!
It happens that "continuous" functions (which are defined by "the limit is equal to the value of the function) have very nice properties so we try to work with continuous functions. But, in fact, one can show that, in a very precise sense, "almost all functions are not continuous". And the "interesting" cases are typically where something is NOT continuous. That's what happens with the derivative. The limit involved in the definition
\lim_{\Delta x\to 0}\frac{f(x+ \Delta x)- f(x)}{\Delta x}
always results in "0/0"! The denominator, \Delta x clearly goes to 0 and then, in order that the limit exist, the numerator f(x+\Delta x)- f(x) must go to 0 (f must be continuous at the point).
What is missing is property of limits that is often under emphasized (or simply ommited):
If, on some neighborhood of a, f(x)= g(x) for all x in the neighborhood except a, then \lim_{x\to a} f(x)= \lim_{x\to a} g(x).
That is, the value of f(x)
at x= a, or whether it even
has a value at x= a, is completely irrelevant to the limit at a.
In particular, in your case, ((x+ \Delta x)- x)/\Delta x= \Delta x/\Delta x. Yes, that is "0/0" and so does not have a value
at \Delta x= 0, but that is irrelevant. For all \Delta x,
except 0, \Delta x/\Delta x= 1 and so the
limit, as \Delta x goes to 0, is 1.
You need to understand that and get used to it, because it will continue to be important! To find the derivative of, say, f(x)= x^2, you would look at the "difference quotient"
\frac{f(x+\Delta x)- f(x)}{\Delta x}= \frac{(x+ \Delta x)^2- x^2}{\Delta x}= \frac{x^2+ 2(\Delta x)x+ (\Delta x)^2- x^2}{\Delta x}
= \frac{2(\Delta x)x+ (\Delta x)^2}{\Delta x}= \frac{\Delta x(2x+ \Delta x}{\Delta x}
Yes, if we just "substitute 0 into [/itex]\Delta x[/itex]", we would get "0/0" which has no meaning. So we
don't do that! Instead we "take the limit as \Delta x goes to 0" by observing that, if \Delta x is NOT 0, we an cancel the \Delta x terms in both numerator and denominator to get 2x+ \Delta x. Since, as long as \Delta x is not 0, those are the same, they have the same limit as \Delta x goes to 0, 2x.
(A latex tutorial for this board is at
https://www.physicsforums.com/showthread.php?p=3977517&posted=1#post3977517)
