# Curvature of a circle approaches zero as radius goes to infinity

1. Apr 15, 2014

### BOAS

Hello,

this isn't a homework problem, so i'm hoping it's okay to post here.

I would like to know the correct way to mathematically express the idea in my title. It is intuitively obvious that as the radius of a circle increases, it's curvature decreases.

I looked it up and found that the curvature of a circle is equal to the reciprocal of it's radius. Certain assumptions are often made when looking at lenses, i.e the wave fronts reaching the lens are parallel, or have 0 curvature - In other words, the object distance is infinitely far away.

But, 1/∞ ≠ 0

So how do I express it properly?

In words, I think it goes something like this - As the radius tends towards infinity, the curvature of the circle tends towards zero.

2. Apr 15, 2014

### Staff: Mentor

Wouldn't you just use the lim 1/r expressions with r-> infinity to express it?

3. Apr 15, 2014

### BOAS

That would be my guess but i'm unsure of how to formulate that.

$lim_{r \rightarrow ∞} \frac{1}{r} = 0$

Like that?

4. Apr 15, 2014

### Staff: Mentor

Yes thats the way I'd express it.

5. Apr 16, 2014

### 7777777

If you imagine a circle with infinite radius, then its circumference is also infinite.
Then what would be the value of pi be? Infinite divided by infinite. Can you say what
it is?
I think the real projective line may be a picture of this kind of "circle":
http://en.wikipedia.org/wiki/Real_projective_line

6. Apr 16, 2014

### Staff: Mentor

The same as always. $\pi$ is a constant (its value never changes).
No. There are several indeterminate forms, including [∞/∞], [0/0], [∞ - ∞], and a few others. These are indeterminate, because you can't determine a value for them.

They usually come up when we are evaluating limits of functions.