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

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

Unparseable latex formula:

f(x,y) = \begin{Bmatrix} \frac{x^2y}{x^2+y^2} & (x,y) \not = (0,0) \\0 & (x,y) = (0,0)



So I did this:

0<x2+y2<δ 0<\sqrt{x^2+y^2}<\delta

∣f(x,y)∣=∣x2yx2+y2∣=∣x2x2+y2∣∣y∣≤∣y∣≤x2+y2 |f(x,y)| = |\frac{x^2y}{x^2+y^2}| = |\frac{x^2}{x^2+y^2}||y| \leq |y| \leq \sqrt{x^2+y^2}

⇒∣f(x,y)∣<δ \Rightarrow |f(x,y)| < \delta

As we need ∣f(x,y)∣<ϵ |f(x,y)| < \epsilon

Do we therefore need δ≤ϵ \delta \leq \epsilon and so you can make delta whatever you want as long as it is less than or equal to epsilon??
Reply 1
Avatar for Palabras
Palabras
2
Correct, and Delta = epsilon is sufficient. (These questions are often done more easily in polar form fyi).
Reply 2
Avatar for adie_raz
adie_raz
OP
1
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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