# Proving a Graphical Limit Analitically

• Aug 11th 2011, 10:17 PM
pflo
Proving a Graphical Limit Analitically
Take a unit square (each side length is 1 unit long). There are 4 sides and the perimeter is 4. Now, take two sides and push them in so that each side becomes two sides, forming a convex hexagon (essentially cutting out a square corner of the original, with each side length 1/2 the length of the original). There are now 6 sides and the perimeter remains 4. Now do this again with the 4 new sides. See the attached picture for a graphical (better) explanation. There are now 10 sides and the perimeter remains 4. Do this again and there will be 18 sides and the perimeter will remain 4.

So, the following sequence is established:

i, n, p
1, 4, 4
2, 10, 4
3, 18, 4
etc…

But when i approaches infinity, n would approach infinity and the picture would approach the triangle (n=3) and the perimeter approaches 2+sqrt(2).

Is this true? How can be illustrated analytically?

Attachment 22035
• Aug 11th 2011, 10:29 PM
Prove It
Re: Proving a Graphical Limit Analitically
All that you're showing is that the perimeter is bounded above by a length of $\displaystyle 4$...
• Aug 12th 2011, 12:03 AM
CaptainBlack
Re: Proving a Graphical Limit Analitically
Quote:

Originally Posted by pflo
Take a unit square (each side length is 1 unit long). There are 4 sides and the perimeter is 4. Now, take two sides and push them in so that each side becomes two sides, forming a convex hexagon (essentially cutting out a square corner of the original, with each side length 1/2 the length of the original).

That hexagon is not convex (which would require that every line line segment connecting two points in the hexagon was inside the hexagon)

CB
• Aug 12th 2011, 12:07 AM
CaptainBlack
Re: Proving a Graphical Limit Analitically
Quote:

Originally Posted by pflo
Take a unit square (each side length is 1 unit long). There are 4 sides and the perimeter is 4. Now, take two sides and push them in so that each side becomes two sides, forming a convex hexagon (essentially cutting out a square corner of the original, with each side length 1/2 the length of the original). There are now 6 sides and the perimeter remains 4. Now do this again with the 4 new sides. See the attached picture for a graphical (better) explanation. There are now 10 sides and the perimeter remains 4. Do this again and there will be 18 sides and the perimeter will remain 4.

So, the following sequence is established:

i, n, p
1, 4, 4
2, 10, 4
3, 18, 4
etc…

But when i approaches infinity, n would approach infinity and the picture would approach the triangle (n=3) and the perimeter approaches 2+sqrt(2).

Is this true? How can be illustrated analytically?

Attachment 22035

The perimiter does not approach $2+\sqrt{2}$ the limit of the sequence of perimiters is 4. This is an example where the properties of the limiting curve is not the limit of the corresponding properties of a sequence of curves.

In fact this is a well known paradox-oid.

CB
• Aug 12th 2011, 05:17 PM
pflo
Re: Proving a Graphical Limit Analitically
Yes, thanks. I was thinking concave but my fingers typed out convex. My bad.