Originally posted by fearlessleader
i still contest.
all your arguments seem to amount to: we can prove that 0/0=anything, so it can't actualy be a number.
but in all the examples you have given, we fill in the discontinuity such that 0/0=1, and no other number.
f(x)=x/x looks like f(x)=1, and we fill it in so that f(0)=1
f(x)=2x/x looks like f(x)=2, and we fill it in so that f( ...[text shortened]... i can think of for which 0/0 can be proven not to = 1 by the argument of removable dicontinuity.
Hearmenow. Let f and g be continuous differentiable functions which both have a zero at x=0. In the following discussion, ''lim...'' means ''limit as x --> 0'' unless otherwise stated. Clearly the function f/g(x) is undefined at x=0 -- it is ''equal'' to 0/0. But
lim f'/g'(x) = lim lim(h-->0) [f(x+h)-f(x)]/[g(x+h)-g(x)]
= lim(h-->0) [f(h)-f(0)]/[g(h)-g(0)]
= lim f/g(x).
When we take limits, we can transform the function in two ways (x becomes y+a and let y-->0 if the original limit was as x --> a or x becomes 1/y and let y-->0 if the original limit was as x--> infinity) which reduce any limit to a limit as the independent variable goes to 0. For any two functions that 'equal' 0/0 at x=0, we can evaluate the limit, if necessary, by evaluating the limit of the derivatives of numerator and denominator. Given this, we can think of functions that 'equal' 0/0 at 0 but have any value we like as x-->0. For example, consider f(x) = (x-tan x)x^-3. This give us 0/0 at x=0, but using the rule proved above, we see that lim f(x) = lim (1-sec^2(x))/3x^2. This still gives us trouble, so we differentiate again and see that lim f(x) = (-2sec^2(x)tan x)/6x. Still a problem, so doing it one more time gives us lim f(x) = -1/3 ''='' 0/0.
BTW, a better way to evaluate limits is to look at the power series expansion of the numerator and denominator, because when x=0 most of the terms drop out and it's quite easy to see what the limit will be, if it exists. This is what motivated the bit about transforming limits to limits as the variable --> 0, along with the fact that we can make the rule used above (called l'Hopital's rule, incidentally) work in more cases.