0/0

based on algebra, we all know that 0/0 is an invalid statment.

but in calculus, it would have to equal one, because in the function
f(x)=x/x
the discontinuity at x=0 is removabel, and we acept that for x=0, f(x)=1, just live every other value of x.

How do you mean the discontinuity is removable?

Originally posted by fearlessleader
based on algebra, we all know that 0/0 is an invalid statment.

but in calculus, it would have to equal one, because in the function
f(x)=x/x
the discontinuity at x=0 is removabel, and we acept that for x=0, f(x)=1, just live every other value of x.

That doesn't mean it equals one, you would define it one. And only in that one case. Because if you would make the function f(x)=2x/x, you would have to define it as 2.

A removable discontinuity is a discontinuity which you can remove by defining a new function in which you explicitly define the point that wasn't thus making the new function continuous.

FL, in the function f(x) = x^2/x, the discontinuity is removable, and 0/0 = 0.

In general, if you want your function to be continuous at a point where pointwise evaluation would have it be undefined, you define its value there to be the limit of the function as the independent variable approches that point. In cases where the limit doesn't exist, the function must have a discontinuity there, no matter how you define its value, because a function is by definition continuous if and only if it takes on the same value at each point as itss limit as the independent variable approaches that point.

For example, sin x/x is continuous if we define its value at x=0 to be 1, because sin x/x --> 1 as x --> 0.

Originally posted by royalchicken
FL, in the function f(x) = x^2/x, the discontinuity is removable, and 0/0 = 0.

Actually, come to think of it, you don't even define 0/0 because it has no meaning. You cannot divide something in no parts. Your object should disappear, catch fire, create a black hole, whatever. So it will remain undefined always. In all these examples you only define f(0), in this case f(0)=0. So your example should be "...is removable and f(0) = 0."

Originally posted by piderman
Actually, come to think of it, you don't even define 0/0 because it has no meaning. You cannot divide something in no parts. Your object should disappear, catch fire, create a black hole, whatever. So it will remain undefined always. In all these examples you only define f(0), in this case f(0)=0. So your example should be "...is removable and f(0) = 0."

Right. I was being a little bit facetious, because it's pretty easy, given a real number a, to think of a function which 'proves' that 0/0=a. We could formally define 0/0 based on the context, but that would be needlessly confusing.

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(0)=2*0/0=2 so 0/0=1
f(x)=x^2/x looks like f(x)=x, and we fill it in so that f(0)=0*0/0=0 in this case, it dosnt matter what 0/0 =, because it is being multiplied by 0, but you get my point.

what about the function f(x)=0/x

it looks like f(x)=0, but if we assum 0/0=1, then it would be discontinus. this would be a removable discontinuity, and we would reset it to f(0)=0. this is the only instance i can think of for which 0/0 can be proven not to = 1 by the argument of removable dicontinuity.

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.

F(x,y) = xy / (x^2 + y^2) for (x, y) not (0, 0)

What do you assign 0/0 here?

Originally posted by TheMaster37
F(x,y) = xy / (x^2 + y^2) for (x, y) not (0, 0)

What do you assign 0/0 here?

i dont see how f(x,y)=0/0 w/out both x and y being 0.


i have to ask knoptfel about gallus nobelus's post, i'll get back to it later.

Mathematics distinguisishes between operations that have no answer at all, e.g. 7/0 , and operations that have a set of possible answers with no convention or other way of working out which answer is correct, e.g 0/0 and 0^0.

An operation which has no valid answer is said to have an answer which is Undefined, whereas an operation which has a number of valid answers is said to have an answer which is Indeterminate.

It is often important to distinguish between these two cases in computer programming, as an Undefined answer is usually a sign that the algorithm is broken, whereas an indeterminate answer is not necessarily a problem, and can often be dealt with by choosing any of the valid answers. Unfortunately all programming languages I have come across will throw the same exception in each case and so the programmer is usually required to identify conditions where an algorithm will return an indeterminate answer and put in special code to set the answer to one of the valid ones without performing the full calculation.

Originally posted by iamatiger
Mathematics distinguisishes between operations that have no answer at all, e.g. 7/0 , and operations that have a set of possible answers with no convention or other way of working out which answer is correct, e.g 0/0 and 0^0.

0/0 has no answer (indetermined) and 0^0 is always 1.

Originally posted by Palynka
0/0 has no answer (indetermined) and 0^0 is always 1.

i agree that 0^0=1, there are no 0s being multiplied, so there is no reason it wouldn't =x^0

Originally posted by royalchicken
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-- ...[text shortened]... t we can make the rule used above (called l'Hopital's rule, incidentally) work in more cases.

but isn't l'hopital's rule used to fine out which converges on infinity or 0 first? so it doesn't actually find out 0/0 or anything. it meerly finds the limit of a function...

Originally posted by genius
but isn't l'hopital's rule used to fine out which converges on infinity or 0 first? so it doesn't actually find out 0/0 or anything. it meerly finds the limit of a function...

True. The WHOLE POINT is that ''0/0'' doesn't mean anything. The point of that post was to show that we can find functions which would give 0/0 at some point but tend to any limit we like as x approaches that point. In that sense, 0/0 is anything we like.