Problem:Give an epsilondelta proof for the existence of the limit, 
> ;
2 2 x  y lim  (x, y) > (0, 0) x + y

> f := (x,y) > (x^2y^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,

in order to show this we factorize the numerator of f(x,y) as, 
> factor(x^2y^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
(xy) is a continuous function at (0,0). But to show that (xy) 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 epsilondelta proof. One more point: If we use the theorem

> ;
2 2 1/2 f(x,y)0 = xy < x+y < (2 (x + y )) = sqrt(2) delta
Ok the last inequality needs a justification. Here it is: 

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?). 

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 epsilondelta hunky punky (i.e. invert the u to get the v... remember?) and get: 
> ;
epsilon delta <  sqrt(2)
Yippiiii....
Mom, I know how to do epsilondelta proofs!!! 