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Limit involving delta-epsilon proof

  1. Nov 1, 2008 #1
    Hi, Everyone,

    Problem asks to prove that limit of x^2 * sin^2 (y) / (x^2 + 2* y^2) as (x,y) approach (0,0) is 0 using delta-epsilon method. Please let me know how to solve, very complicated.

    Thanks.
     
  2. jcsd
  3. Nov 1, 2008 #2

    gabbagabbahey

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    What have you tried so far?
     
  4. Nov 1, 2008 #3
    What I tried so far is here.
    [​IMG]
     
  5. Nov 1, 2008 #4

    gabbagabbahey

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    The condition that you need to satisfy is that

    [tex]|f(x,y)-0|=\left| \frac{x^2sin^2(y)}{x^2+2y^2} \right| \leq \varepsilon[/tex]

    So showing that [itex]x^2sin^2(y) \leq \varepsilon[/itex] is insufficient to prove that the limit is zero. (I'm assuming that [itex]xsin(y)[/itex] was a typo)

    Start by using the fact that [itex]0 \leq sin^2(y) \leq 1[/itex] and then find some function of delta [itex]u(\delta)<\varepsilon[/itex] such that

    [tex]\left| \frac{x^2sin^2(y)}{x^2+2y^2} \right| \leq u(\delta)[/tex]
     
  6. Nov 2, 2008 #5
    Thanks a lot. Check it out. Here is deal. Let me know if it good.
    [​IMG]
     
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