# Thread: How to prove that this limit does exists...

1. ## How to prove that this limit does exists...

How can I correctly justify, demonstrate, and properly determine that this limit exists?

lim (x,y) --> (0,0) $\sqrt{x^2+y^2}$

2. Originally Posted by HypatiaDaVinci
How can I correctly justify, demonstrate, and properly determine that this limit exists?
lim (x,y) --> (0,0) $\sqrt{x^2+y^2}$
The distance .

So

3. Thank you very much for the answer. How do you get to the conclusion that $\varepsilon=\delta$ I mean... how do you directly relate them?

4. Originally Posted by HypatiaDaVinci
Thank you very much for the answer. How do you get to the conclusion that $\varepsilon=\delta$ I mean... how do you directly relate them?
To prove this by the definition, you should state that
"For all x,y in R, for all epsilon > 0, there exists a delta > 0 such that [d(x, y) < delta] implies [sqrt(x^2 + y^2) < epsilon]
since "d(x, y)" and "sqrt(x^2 + y^2)" are the same thing, you can choose delta to be the epsilon that is given.

5. Thank you very much, I think I got it now