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

 P: 273 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.
 P: 3,097 Wouldn't you just use the lim 1/r expressions with r-> infinity to express it?
P: 273
 Quote by jedishrfu Wouldn't you just use the lim 1/r expressions with r-> infinity to express it?
That would be my guess but i'm unsure of how to formulate that.

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

Like that?

 P: 3,097 Curvature of a circle approaches zero as radius goes to infinity Yes thats the way I'd express it.
 P: 21 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
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P: 21,399
 Quote by 7777777 If you imagine a circle with infinite radius, then its circumference is also infinite. Then what would be the value of pi be?
The same as always. ##\pi## is a constant (its value never changes).
 Quote by 7777777 Infinite divided by infinite. Can you say what it is?
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.
 Quote by 7777777 I think the real projective line may be a picture of this kind of "circle": http://en.wikipedia.org/wiki/Real_projective_line

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