Limit epsilon/delta proof

**Roam** · May 26th 2009, 11:13 PM

Show that the limit as (x,y) -> 0 doesn't exist.

Attempt:

To prove it, I must use the definition of limit (the epsilon and delta thing).

Well, if the limit exists, and is 1, given any $\epsilon >0$ we can find a $\delta >0$ such that $\sqrt{x^2 + y^2} < \delta$ if $|f(x,y) -1| < \epsilon$

perhaps we can ty $\epsilon = 1/2$

I'm not sure how to complete this proof to conclude that $\lim_{(x,y) \to (0,0)} f(x,y)$ doesn't exist... some help with this is very much appreciated.

**Roam** · May 27th 2009, 02:56 AM

Am I on the right track? ... I need some help finishing this proof.

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