Name: Jason

Who is asking: Student

Level: Secondary

Question:

What do you get when dividing zero by infinity? Our Calculus teacher was pretty sure that the expression was indeterminate from. However, if this is so...Why? Zero divded by any number (except zero) is zero, true. Any number (except infinite) over infinite is zero. So, why isn't Zero divided by infinite zero. A simpler way if I had 4 potatoes and was to split them among 2 friends, each friend would get 2 potatoes. However, if I had 0 potatoes and split them a infinite number of ways, each person would still have 0. Explain please!

Hi Jason,
I like your potato example.

It bothers me somewhat that we talk about infinity as if it were a number that we can treat just as we treat other numbers. We all do it, but when faced with this kind of question it is important to ask where the infinity (and the zero) comes from. I am going to use function notation as it makes it easier to say what I want.

I assume you have a fraction of the form ^{f(x)}/_{g(x)} , and that as x approaches a, f(x) approaches zero and g(x) approaches infinity. The question is then, what is the limit of ^{f(x)}/_{g(x)} as x approaches a? To make my life a little easier I am going to assume that f(x) and g(x) are never negative. That way I don't have to deal with negative signs or absolute values.

If f(x) = 0 for every x then^{f(x)}/_{g(x)} = 0 for every x, and hence the limit is zero. (This is your potato example.) Since g(x) approaches infinity as x approaches a, as x gets close to a, g(x) > 1. Thus as x gets close to a,

0 < ^{f(x)}/_{g(x)} < f(x).
Hence ^{f(x)}/_{g(x)} gets squeezed between 0 and f(x), and f(x) is approaching zero. Thus ^{f(x)}/_{g(x)} must also approach zero as x approaches a.
If this is what you mean by "dividing zero by infinity" then it is not indeterminate, it is zero.

Penny