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

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

     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<\sqrt{x^2+y^2}<\delta

     |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}

     \Rightarrow |f(x,y)| < \delta

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

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