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Exploring the Distance from (0,0) to (1,1) with Limits

Date: 10/15/2006 at 00:28:21
From: Marc
Subject: Does the distance from (0,0) to (1,1) have a discontinuity

Any monotonic route from (0,0) to (1,1) along a path of only right 
angles will have a total distance of 2.  As the number of "steps" 
increases to infinity, the distance is constant.  Yet in the limit, 
the distance traveled should converge to the diagonal length of sqrt
(2).  How can this be?

I'm confused because my intuition tells me this function has got to be 
continuous.  I don't know what reasoning to bridge the gap between a 
right angled path and the diagonal route.

I understand arc length derivation, but that also makes the leap from 
right-angled steps to diagonal in the differential [ds = sqrt
(dx^2 + dy^2)].  Why can't we press onto smaller and smaller steps and 
still get the answer sqrt(2)?

If I take n equal length steps, the distance traveled is n*(1/n)*2.  
The limit of this is 2*inf/inf -LHop-> 2*1/1 = 2.  But as n increases, 
I'm getting arbitrarily close to walking the diagonal, which I know 
has length sqrt(2).  Why doesn't my limit calculation apply here?



Date: 10/17/2006 at 15:55:07
From: Doctor Rick
Subject: Re: Does the distance from (0,0) to (1,1) have a discontinuity

Hi, Marc, thanks for writing to Ask Dr. Math.

Your intuition goes astray not on the continuity issue, but when you 
assume that in the limit, the distance should converge to the length 
of the diagonal line.  Here is how I answered a similar question a 
year ago:

=================================================================
Question: 

Hi, my math teacher one day said it was ridiculous to do this.

Take a triangle with both of the smaller side (a and b) equal to one. 

Using Pythagoras it is easy to see that the other side is (c) equal to 
square root of 2.  

Ex:  
  | \
1 |  \ square of 2
  |___\
    1


Why then can't you approximate the last line by doing a small 
escalator that if pushed to infinity it would approximate one doing 
something like this 

   __
1 |  |
  |   --
  |_____|  then split again  
     1
   _
  | |
  |  -
1 |   |
  |    -
  |_____|
     1

if you continue like this how can you say that adding the little lines 
you won't get a good approximation of the line c??  

To put it into a reasonable mathematical inequality such as 

  lim x + lim x  ain't equal to ... .. 

I'm pretty sure it has to do with limits and infinite sums but I can't 
put my finger on it... perhaps there's an impossible inversion?  

Answer:

You are assuming that, in the limit, the total length of that "zigzag" 
line is equal to the length of the straight line joining its 
endpoints.  This is not true.

Consider the length of a "zigzag" before you take the limit:

  +---+
    \ |
      +---+
        \ |
          +---+
            \ |
              +---+
                \ |
                  +

In each of the small triangles, the ratio of the sum of the horizontal 
and vertical legs to the hypotenuse is 2:sqrt(2), or sqrt(2):1.  Thus 
the ratio of the total length of the zigzag to the length of the 
straight line is also sqrt(2):1.  Thus if the length of the straight 
line is sqrt(2), the length of the zigzag is 2.

This is true no matter how small you make the triangles!  The sequence 
of lengths of the zigzags is 2, 2, 2, 2, 2, ... What is the limit of 
this sequence?  Obviously it is 2, not sqrt(2).

The zigzag, in the limit, may "look" like a straight line--but it is 
not a straight line!  Its length is not equal to the length of the 
straight line.  This is not intuitive.  The message is: Don't judge 
based on the appearance of the limiting case.  Examine the sequence 
carefully *before* you take the limit.
=================================================================

Does that answer your question?

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Calculus
High School Triangles and Other Polygons

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