Distance formula

[Tolnaftate2004]

Originally Posted by Kristi

It doesn't matter what your calculator says, 0/0 is 0

I'm not using a calculator.

So,
lim x->0 of 0/x = 0
lim x->0+ of x/0 = infinity
lim x->0- of x/0 = -infinity
lim x->0 of x/x = 1
lim x->0 of (x/2)/x = 1/2
lim x->0 of nx/x = n (n!=0)

I know how to show 0 = 0, but how about the rest of these? How is infinity = 0? If n != 0, how is it 0, then? 0^0 or 0/0 cannot simply be described with one number.

An additional note:
lim x-> 0 of 0^x = 0
lim x-> 0 of x^0 = 1

All "modern algebra" does is lead me to believe that 0/0 is some set that includes {0,1} at least.

[Angel_Light]

My brain hurts now.

[Kristi]

Originally Posted by Tolnaftate2004

I'm not using a calculator.

So,
lim x->0 of 0/x = 0
lim x->0+ of x/0 = infinity
lim x->0- of x/0 = -infinity
lim x->0 of x/x = 1
lim x->0 of (x/2)/x = 1/2
lim x->0 of nx/x = n (n!=0)

I know how to show 0 = 0, but how about the rest of these? How is infinity = 0? If n != 0, how is it 0, then? 0^0 or 0/0 cannot simply be described with one number.

An additional note:
lim x-> 0 of 0^x = 0
lim x-> 0 of x^0 = 1

You are using limits to describe finite concepts. We already know 0 can be divided into 0 parts. Limit definitions are solid when finite definitions do not exist, eg diving by zero. Generally speaking, you apply the limit to the modifying operation, eg division, raising to a power, etc.

[Tolnaftate2004]

Originally Posted by Kristi

You are using limits to describe finite concepts.

A limit can be applied anywhere. A limit only exists when the limit is finite, anyway. This statement really confuses me.

Originally Posted by Kristi

Limit definitions are solid when finite definitions do not exist, eg diving by zero.

But you keep saying 0/0 = 0. 0 is neither -infinity nor infinity, so it is finite.

Originally Posted by Kristi

Generally speaking, you apply the limit to the modifying operation, eg division, raising to a power, etc.

Each of the limits above are valid for evaluating 0/0. The "modifying operation" is rather ambiguous in this case. 0/0 could be split up to 0*1/0, or any other number of ways and we'd have a few operations to choose from.
Depending on how we approach 0/0, we can get infinitely different answers. Consider the equation (3x^2+nx)/x. At x=0, we have 0/0, and there is a removable discontinuity, but we can still evaluate the limit and it gives us n.

The function happens to be continuous. If we say that 0/0 = 0, we've broken the continuity.