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Using the Delta-epsilon definition but for two variables?

  1. Sep 10, 2007 #1
    I have probably over thought the whole thing... but I can't seem to find any place to start with this one:

    Using the formal definition of a limit:

    f(x,y)= y / (x^2 + 1) e=0.05
  2. jcsd
  3. Sep 10, 2007 #2

    D H

    Staff: Mentor

    For multiple dimensions you need to extend the epsilon-delta notation to incorporate the epsilon-neighborhood of the point p0 = (x,y), which is the set of all points p such that the distance between p0 and p is less than epsilon.
  4. Sep 10, 2007 #3
    The limit seems to be 0.
    So if [tex]0<|y|<\delta[/tex] and [tex]0<|x|<\delta[/tex].

    [tex]\left| \frac{y}{x^2+1} \right| \leq \frac{|y|}{x^2} < \frac{\delta}{x^2}[/tex].

    Now use single variable analysis that [tex]\frac{\delta}{x^2}[/tex] can be made sufficiently small.
  5. Sep 11, 2007 #4


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    Staff Emeritus
    Science Advisor

    limit as (x,y) goes to what?
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