Curvature of a circle approaches zero as radius goes to infinity

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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.
 
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Wouldn't you just use the lim 1/r expressions with r-> infinity to express it?
 
jedishrfu said:
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?
 
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
 
7777777 said:
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).
7777777 said:
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.
7777777 said:
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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