Euler lived in the 1700s. Not sure how that would have affected your training unless you’re much older than average.
Cauchy and Taylor both formalized many concepts, again in the 18th and 19th century.
What are you thinking is ad box?
Cauchy and Taylor both formalized many concepts, again in the 18th and 19th century.
What are you thinking is ad box?
limit as h -> 0 of (f(x+h) - f(x)) / h.
That's well-defined on (0, inf) but not on [0, inf). So you can't just evaluate at h=0 and be done with it.
The intuition is 'as h gets smaller and smaller, the ratio gets closer and closer to a new function of x'. But many high-school students aren't given a clear definition of what it means for one function to be 'close' to another, or what it means for x to 'get smaller and smaller'.
To see the confusion more clearly, try having the debate about whether 0.999... = 1 with someone who doesn't understand what a limit is.
I think your proof goes wrong since you haven't justified how arithemtic operations work with infinite decimals. AFAIK the only way to add non-terminating decimals is to convert them to fractions (or sequences of fractions as with pi, e, etc), add the fractions, and convert them back. So if you convert 0.999... and 9.999... to fractions, you've assumed the conclusion.
To play devil's advocate, I can try to rephrase your proof without infinite decimal arithmetic as follows.
Assume
n = 0.999... = 1 - epsilon, where epsilon is 'infinitesimal' (an ill-defined version of not-quite-zero). We'd like to show that epsilon is zero.
10n = 9.999 = 10 - 10epsilon
9n = 9.999 - (1 - epsilon) = 9 - 9epsilon
9n = 8.999 + epsilon = 9 - 9epsilon
The only way to get the epsilons to cancel is to assume epsilon = 0, which is to assume the conclusion.
Any digit over 9 equals a decimal of zero point itself repeating. I.E.
1/9 = 0.111... 2/9 = 0.222... 3/9 = 1/3 = 0.333... ... 1 = 9/9 = 0.999...