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Choosing delta in a Delta-Epsilon Proof of limits in more than one dimension watch

1. I need to prove that the following function is continuous at (0,0):

So I did this:

As we need

Do we therefore need and so you can make delta whatever you want as long as it is less than or equal to epsilon??
2. Correct, and Delta = epsilon is sufficient. (These questions are often done more easily in polar form fyi).
3. (Original post by Palabras)
Correct, and Delta = epsilon is sufficient. (These questions are often done more easily in polar form fyi).
So I can assign any arbitrary delta I want as long as it is less than or equal to epsilon. To prove it do I actually have do do this?? For example, is what I have sufficient, or when I am done, do I have to say "Therefore let delta equal epsilon" and thus the function is continuous.

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Updated: March 29, 2011
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