TITLE: An epsilon-delta proof of a limit
# !b2 Problem:
Give an epsilon-delta proof for the existence of the limit,
> ;
                                              2    2
                                             x  - y
                                 lim         -------
                           (x, y) -> (0, 0)   x + y

# !b3 Solution:
> f := (x,y) -> (x^2-y^2)/(x+y):
# First notice that if the limit exists it must be the value
obtained when we compute it along any path aproaching (0,0).
So let us take the simple path y=x as x->0. We have,
> ;
                                       2    2
                                      x  - x
                               lim    ------- = 0
                              x -> 0   x + x

# We conclude that the only candidate for the limit is 0.
For the limit to be zero we must show that,
!c0 for any epsilon > 0 there is delta > 0 such that
!c0 when 0 < |(x,y)-(0,0)| < delta  we have |f(x,y)-0| < epsilon
# in order to show this we factorize the numerator of f(x,y) as,
> factor(x^2-y^2);
                                (x - y) (x + y)

# and then for all (x,y) different from (0,0) we have
> 'f(x,y)' = simplify(f(x,y));
                                f(x, y) = x - y

# At this point we can be certain that the limit exists and that it
is in fact 0. We could see it, that is, if we accept the fact that 
(x-y) is a continuous function at (0,0). But to show that (x-y) is
continuous you need to show the existence of the limit so the 
complete argument is circular. There is no way but to use
the epsilon-delta proof. One more point: If we use the theorem

!c0 (limit of a sum) = (sum of the limits)
the epsilon-delta proof would be much easier... but why making it
easy if we can make it more complicated? By the way the real practical
value of the epsilon-delta definition is that it allows to show 
theorems like the one above. OK back to the proof.
We need to show that, 
!c0 |f(x,y)-0| < u(delta) = (some function of delta)
and then find a delta such that
!c0 u(delta) < epsilon
if u() is increasing with inverse v() then v() is also increasing
and by applying v on both sides of the inequality we obtain,
!c0 delta < v(epsilon)
and we are DONE! That's the usual plan for epsilon-delta proofs.
That's at least the theory. Now back to the dirty specifics of 
the real (mathematical) world
> ;
                                        2    2  1/2
    |f(x,y)-0| = |x-y| < |x|+|y| < (2 (x  + y ))   = sqrt(2) delta

# Ok the last inequality needs a justification. Here it is:
# !b3 Lemma:
For all (x,y) we have,
> ;
                2    2  1/2
 |x|+|y| < (2 (x  + y ))   
 
# the <'s are really less or equal in this page (OK?).
# !b4 Proof:
# here is the trick,
> ;
                           2        2                        2
       0 <  (| x | - | y |)  = | x |  - 2 | x | | y | + | y |

# from where we deduce that
> ;
                                           2        2
                          2 | x y | < | x |  + | y |

# Put this on the ONE hand. On the OTHER hand put
> ;
                             2        2                    2
              (| x | + | y |)  = | x |  + 2 | x y | + | y |

# and use the stuff you got on the ONE hand to show that
> ;
                                   2           2        2
                    (| x | + | y |)  < 2 (| x |  + | y | )

# Finally use the fact that sqrt() is an increasing function and
take sqrt() on both sides of this last inequality and we are
done with the Lemma!
# With the Lemma under the belt we do the final epsilon-delta hunky punky
(i.e. invert the u to get the v... remember?) and get:
> ;
                           epsilon
                   delta < -------
                           sqrt(2)
# Yippiiii.... 
!b2 Mom, I know how to do epsilon-delta proofs!!!


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