Is 0/0 the Key to Understanding Infinity?

[Mentallic]

Both approach towards zero so they will both tend to be two very small numbers. This points to a similarity. They are not arbitrary, they are specifically very small. Hence it comes to reason that while the result can be anything between -∞ and +∞, they will have a bias towards 1 rather than ∞.

If you're happy to avoid any Mathematical rigour and just throw ideas out there, then I'll have my take on it too.

I could also argue that since both numbers are tending to 0, and "many" indeterminate forms of 0/0 are far from 1, being either $\pm\infty$ then this is evidence that points towards it having a bias towards infinity rather than 1. I'm imagining that the only reason you're seeing a bias towards 1 is because x/x=1 for every other real value of x. Or, how about we take another approach? If we consider the limits that are non-negative, then the value for 0/0 must be $[0,\infty)$ and since if we consider

$$\lim_{(a,b)\to 0}\frac{a}{b}$$

then the value is only equal to 1 when a approaches 0 at the same rate as b, so there are going to be equally many values of a that approach zero faster than b (giving a value of < 1) as there are going to be values of a approaching 0 slower than b (giving a value > 1), so does this mean 1 is equidistant between 0 and $\infty$?

*** There is of course no mathematical rigour in these statement I've made, and I absolutely do not stand by them.

 

[Mark44]

Is it not possible to produce a fuzzy output?

No, and you have been told this repeatedly. Division, when it is defined, produces a single result. Furthermore, division by zero is undefined.

My understanding is that a/b with them independently approaching zero, will produce a fuzzy 'anything' number around 1.

Then your understanding is flawed.
These three limits are all of the [0/0] indeterminate form, but the limit values are wildly different.
$$\lim_{x \to 0} \frac{x}{x^2} \text{does not exist} $$
$$\lim_{x \to 0} \frac{x^2}{x} = 0$$
$$\lim_{x \to 0} \frac{x}{x} = 1$$

What you seem to be missing is that even though both numerator and denominator are approaching zero, how quickly one or the other is approaching zero is the determining factor.

PS. If they are equal it will produce 1.

If they both approach zero at the same rate, the limit will be zero.

A weird postulation this may produce is that if that 'anything' has a tendency to be closer to 1 rather that infinities then it might point towards why physical numbers tend to not be infinite.

Nonsense.

Both approach towards zero so they will both tend to be two very small numbers.

Yes, of course. That's what "approaching zero" means, but again, what's important is how quickly one or the other (or both) are approaching zero.

This points to a similarity. They are not arbitrary, they are specifically very small. Hence it comes to reason that while the result can be anything between -∞ and +∞, they will have a bias towards 1 rather than ∞.

No.

The same would be true for ∞/∞.

No, absolutely not, and for the same reason I gave above. The important consideration is not that both numbers are getting arbitrarily large, but rather, how quickly one or the other (or both) is getting large.

 

[Nugatory]

Both approach towards zero so they will both tend to be two very small numbers. This points to a similarity. They are not arbitrary, they are specifically very small. Hence it comes to reason that while the result can be anything between -∞ and +∞, they will have a bias towards 1 rather than ∞.

Here is one specific and very important (it's the basis for all calculus, and it's been known and well understood since the 17th century) example:$$\lim_{a\to0}\frac{f(x+a)-f(x)}{a}$$

It should be clear that both the numerator and the denominator are approaching zero. However, the value of this expression doesn't have a "bias towards 1" - it is the slope of the graph of $f$ at the point $x$ (the $a$ disappears everywhere when you take the limit) and it's only going to have a "bias towards 1" if $f$ is specifically the function $f(x)=x$. Trivially, it has the value $A$ when $f$ is the function $f(x)=Ax$ and there's no reason why $A$ should be anywhere near one. It gets even more interesting and even less biased towards one if $f$ is a more interesting function (try it with $f(x)=2x^2$ as an exercise).

As I said, this application of the $0/0$ machinery was discovered along with differential calculus in the 17th century. Until you've worked through it, you're restricting your mathematical understanding to the state of the art - four centuries ago.

 

[micromass]

I think it's time to close this. Many posters have given clear and correct answers. It's up to the OP to decide whether he wants to listen or not.