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Curvature of a circle approaches zero as radius goes to infinity

  1. Apr 15, 2014 #1
    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. jcsd
  3. Apr 15, 2014 #2

    jedishrfu

    Staff: Mentor

    Wouldn't you just use the lim 1/r expressions with r-> infinity to express it?
     
  4. Apr 15, 2014 #3
    That would be my guess but i'm unsure of how to formulate that.

    [itex]lim_{r \rightarrow ∞} \frac{1}{r} = 0[/itex]

    Like that?
     
  5. Apr 15, 2014 #4

    jedishrfu

    Staff: Mentor

    Yes thats the way I'd express it.
     
  6. Apr 16, 2014 #5
    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
     
  7. Apr 16, 2014 #6

    Mark44

    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.
     
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