what is 0/0?

Question

what is the value of 0/0?is it infinite or undetermined?

Answer 1

In the expression x/x, as x approaches 0, the limit is 1; but in the expression x2/x, as x approaches 0, the limit is 0. In each case, however, if the limits of the numerator and denominator are evaluated and plugged into the division operation, the solution is 0/0. So (roughly speaking) 0/0 might mean 1, or it might mean 0 (and in fact, by appropriate examples, it can be made to mean anything); so it is indeterminate.

Note that an indeterminate form is not the same thing as an expression that is undefined. Although most indeterminate forms (such as 0/0) are also undefined, the indeterminate form 00 is usually defined to be 1 (see Empty product). On the other hand, the expression 1/0 is undefined (at least as a real number), but it is not an indeterminate form (because such an expression, such as 1/x as x approaches 0, consistently diverges to infinity in some way).

Answer 2

0/0 is undefined. Period. The reason is simple. If 0/0 had a value (call it x) then

0/0 = x implies that 0 = 0x which is true for *all* x.

The argument that the limit of x/x as x-->0 is 1 has a certain validity, but people who claim that somehow 'proves' that 0/0 = 1 simply show their ignorance of what is meant by a limit. The function x/x is simply equal to 1 for all x *except* at x = 0 where it is undefined. It literally has a 'hole' in the line y = 1. And that hole has the dimension of a single point.

In Calculus, the 'derivative' of a function f(x) for some value of x is defined to be

f'(x) = (f(x+h) - f(x))/h in the limit as h-->0

This is evaluated by evaluating the denominator and finding some way to 'factor out' an h term so that it may be divided out (not 'canceled', I *hate* that term because it implies that the h's somehow magically 'go away'. They don't )

An example:

f(x) = x² so f'(x) = (f(x+h) - f(x))/h as h-->0 then

((x+h)² - x²)/h = (x²+2xh+h²-x²)/h = (2xh+h²)/h = h(2x+h)/h

and we are left (after dividing out the h's) with

f'(x) = 2x + h

now, if h-->0 the value of this 'derived' function goes to 2x and we say

f(x) = x² ==> f'(x) = 2x

The analogy I used to use (when I tought undergrad Calculus) is that of a garden path that's paved with little round 'stepping stones'. If you come to a place where a stone is missing, you still know where the stone would be if it *was* there.

And it's the exact same thing with limits. The actual point on the function (f(x+h) - f(x))/h at h = 0 simply does not exist. But, if it *did* exist, in the example above it would be at 2x.

Consider the simple line y = x+2. Is the point (1,3) on this line? Sure it is. Now consider

y = (x²+x-2)/(x-1) = (x-1)(x+2)/(x-1) = x+2

Is the point (1,3) contained in this function?

**NO**, it is not. And the reason is that the division operation used to make the (x-1) term in the numerator and denominator 'go away' is valid if, and only if, x is not equal to 1.

Again, there is a 'hole' in the line at (1,3)

Hope all of that helps.


Doug

Answer 3

1

Answer 4

It is not infinite but it is indeterminite. Reason being... 0/0=1,0/0=2, 0/0= 3 and so on so you cannot determine which is the right answer

Answer 5

0/0 is undefined. You can never have a 0 in the denominator. If you do, the world blows up and we all die. end of story.

Answer 6

2/0 = 0
0/0=0
anything divided by 0 is 0

Answer 7

Imagine you have 0 cookies and you split them equally among 0 friends.See it doesn't make sense and cookie monster is sad that there are no cookies and you are sad that you have no friends.

Answer 8

i imagine of it really is purely a glitch. i have were given both watched and uploaded videos which have had 0:00 as there time and they have worked stunning. you're in a position to continuously delete the video and upload it back.

Answer 9

OK here's how it goes:
0/x= 0
x/0= infinity
0/0= undefined

Answer 10

anything divided by 0 is 0.
eg: 2/0 = 0
even 0/3 = 0
so, the end ans is 0.