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May 28th, 2015, 10:04 AM   #11
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 Originally Posted by lemgruber I refer the statement what a straight is a curve with radius tends to infinite
Let's review what this means. You have a series of circles where the radius of the individual circles is increasing without bound. Perhaps the first circle has radius 1, the second radius 2, the third radius 3, and so on.

You could look at the curvature of each circle, for example, and find that it is 1 for the first, 1/2 for the second, 1/3 for the third, and so on. You could then note that the curvature of the sequence tends toward a limit, which is 0. This is what is means (formally) when a person says that as the radius increases without bound ("tends to $\infty$") that the curvature tends to 0.

What claim are you making about this sequence of circles? If you're defining a function f(C) is 1 if C is a curve and 0 otherwise, then f of the first circle is 1, f of the second circle is 1, and so on; the limit is, of course, 1.

 May 28th, 2015, 12:15 PM #12 Senior Member   Joined: Dec 2014 From: Brazil Posts: 203 Thanks: 1 I say what the function when radius tends to ∞, not be defined in this point, that's all. Lucio
 May 28th, 2015, 01:09 PM #13 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2643 Math Focus: Mainly analysis and algebra It doesn't have to be defined at the limit point for the limit to exist.
 May 28th, 2015, 01:26 PM #14 Senior Member   Joined: Dec 2014 From: Brazil Posts: 203 Thanks: 1 Forgive-me the error , I would to say the function not be defined when the radius tends to infinite, I wanna say it's limit doesn't exists what this occurs. What I say about the point, do not exist such thing like infinite point. Lucio Last edited by lemgruber; May 28th, 2015 at 01:28 PM.
 May 28th, 2015, 02:13 PM #15 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2643 Math Focus: Mainly analysis and algebra Well clearly the function is well defined for all finite radius. And in these cases the curvature is non-zero (positive, in fact). In the limit the curvature is zero, but the function is clearly 1. It doesn't work to claim that the limit doesn't exist just because it doesn't say what you want it to say.
 May 28th, 2015, 02:20 PM #16 Senior Member   Joined: Dec 2014 From: Brazil Posts: 203 Thanks: 1 I agree. Lucio

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