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BASIC limit

  1. Oct 27, 2012 #1
    1. The problem statement, all variables and given/known data

    show that $$f(x,y) =y+x^2cosy $$ has a limit 0 at (0,0) by the ε-δ definition.

    2. Relevant equations



    3. The attempt at a solution
    $$|y+x^2cosy| ≤ |y|+|x^2|$$ (by tri. inequ. and $$|cosy|≤1$$
    then can I suppose $$|x^2|<|x|$$ , since $$|x|<1$$,
    then $$|y+x^2cosy| ≤ |x|+|y| ≤ 2\sqrt{x^2+y^2} $$ ?
    if not , how can I do it by the ε-δ definition
     
  2. jcsd
  3. Oct 27, 2012 #2
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  4. Oct 27, 2012 #3

    HallsofIvy

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    Given [itex]\epsilon> 0[/itex], show that there exist [itex]\delta> 0[/itex] such that if [itex]\sqrt{x^2+ y^2}< \delta[/itex] (this is the distance from (x, y) to (0, 0)) then [itex]|y+ x^2cos(y)|<\epsilon[/itex].

    From your last inequality, [itex]|y+ x^2cos(y)|< 2\sqrt{x^2+ y^2}[/itex] you just need to choose [itex]\delta< \epsilon/2[/itex].
     
  5. Oct 27, 2012 #4
    thanks you guys, actually I would want to know can I suppose x^2|<|x| , since |x|<1
     
  6. Oct 28, 2012 #5

    Zondrina

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    Homework Helper

    Hmm depends on how small you want your neighborhood to be in terms of your choice of δ.

    Otherwise you can assume that |x| ≤ (x2+y2)1/2 < δ and |y| ≤ (x2+y2)1/2 < δ

    So that : [itex]|y| + |x|^2 < δ + δ^2 = δ(δ+1)[/itex] Then for δ≤1, we would have :

    [itex]|y| + |x|^2 < δ + δ^2 = δ(δ+1) ≤ 2δ ≤ ε[/itex]

    So as long as δ ≤ min{1, ε/2} you're good.
     
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