Epsilon-Delta Proofs of Limits

[GluonZ]

Trying to press on through Epsilon-Delta proofs of limits (for more than one variable) and yet there's only one example I've found thus far of even a multi-variable Epsilon-Delta proof.

Would it be possible for someone to solve the Epsilon-Delta proof of the limit:

(xy^2)/(x^2+y^2). Note: The limit does *not* exist... along straight-line paths the limit is 0, yet on the x=y^2 parabola the limit is 1/2.

x=y^2

IE: (y^2)y^2/(y^2)^2+y^4 = y^4/2y^4 = 1/2.

Profs always say Epsilon-Delta proofs are among the hardest things to get in math... go figure! You try and get examples and its always the same 3 recycled over and over, and all are examples of limits which exist. Most are even polynomial aswell, which makes it super-easy. (Except you never see Polynomial proofs on exams, no they must give you rationals)

 

[matt grime]

But you don't want to use epsilon delta *proofs* for this. You'd use those to show the limit did exist and was something. Not that you've said what x and y are tending to in the limit, by the way.

 

[GluonZ]

Whoops, thanks. X and Y are tending to (0,0)...

I agree it'd be stupid to use Epsilon-Delta proofs for this, there's already a contradiction evident (lim (x,y)->(0,0) through x=y^2 = 1/2... not 0.) Just was hoping that provided Epsilon-Deltas prove a limit exists, I could find out what an Epsilon-Delta proof would look like if it was unable to prove the limit (being that the limit did not exist).

 

[StatusX]

The limit of f(x) as x->a is L iff for every e>0, there is a d>0 so |x-a|<d implies |f(x)-L|<e (roughly speaking). Then negating this, you get that the limit is not L if there is some e>0 (ie, e is fixed throughout the rest of this), such that for all d>0 there is some x with |x-a|<d but |f(x)-L|>=e. Do you understand how this is the negation of the above? So to prove a function has no limit, you just need to show that every number is not its limit. With a little work you should be able to prove that this is the case whenever the limit is different along different paths.

 

[GluonZ]

Ah, that makes sense. Thanks guys.