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

• April 25th 2011, 01:44 PM
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) $\sqrt{x^2+y^2}$
• April 25th 2011, 02:05 PM
Plato
Quote:

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 http://quicklatex.com/cache3/ql_8337...4a22498_l3.png.

So http://quicklatex.com/cache3/ql_9539...952daeb_l3.png
• April 25th 2011, 07:17 PM
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?
• April 25th 2011, 07:21 PM
TheChaz
Quote:

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.
• April 26th 2011, 08:51 AM