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Homework Help: Limit of a function in two variables

  1. Jun 26, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove with [itex]\epsilon-\delta[/itex]: [itex]Lim_{(a,b)\rightarrow(0,0)}\frac{sin^{2}(a-b)}{\left|a\right|+\left|b\right|}=0[/itex]
    Hint: [itex]\left|sin(a+b)\right|\leq\left|a+b\right|\leq\left|a\right|+\left|b\right|[/itex]

    2. Relevant equations

    [itex]0<\sqrt{(x-x_{0})^{2}+(y-y_{0})^{2}}<\delta[/itex]
    and
    [itex]\left|f(x,y)-L\right|<\epsilon[/itex]

    3. The attempt at a solution

    I tried the polar conversion, which was just messy and got me nowhere. Then I tried inputting for [itex]\epsilon-\delta[/itex] formula: [itex]0<\sqrt{x^{2}+y^{2}}<\delta[/itex] and [itex]\frac{sin^{2}(a-b)}{\left|a\right|+\left|b\right|}-0<\epsilon[/itex] so that [itex]sin^{2}(a-b)<(\left|a\right|+\left|b\right|)\epsilon[/itex]
    The tutorials I've seen use the [itex]\epsilon[/itex] inequality and transform it into the [itex]\delta[/itex] inequality, but I don't know how to do this. I'm just looking for a way to solve it, not necessarily using the "hint".
     
    Last edited: Jun 26, 2012
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  3. Jun 26, 2012 #2

    SammyS

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    The hint appears to be very useful here.

    If [itex]\left|\sin(a+b)\right|\leq\left|a+b\right|\leq \left|a\right|+\left|b\right|[/itex],

    then [itex]\left|\sin(a-b)\right|[/itex]
    [itex]=\left|\sin(a+(-b))\right|\leq\left|a+(-b)\right|\leq\left|a\right|+\left|-b\right|[/itex]
                   [itex]=\left|a\right|+\left|b\right|[/itex]​
     
  4. Jun 26, 2012 #3
    Okay, that makes more sense. So if [itex]sin^{2}(a-b)\leq sin(a-b)\leq\left|a\right|+\left|b\right|[/itex]
    Then
    [itex]\left|a\right|+\left|b\right|<\left|a\right|+\left|b\right|\epsilon[/itex]
    So
    [itex]1\leq \epsilon[/itex]

    However, I've been taught to solve for [itex]\delta[/itex] first, and then prove with Given [itex]\epsilon>0[/itex] and Choose [itex]\delta= constant*\epsilon[/itex]. Now if I have [itex]\epsilon[/itex] first, how do I construct a proof this way ?
     
  5. Jun 26, 2012 #4
    Any guidance is appreciated :)
     
  6. Jun 26, 2012 #5

    SammyS

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    That basically shows that [itex]\displaystyle \frac{\sin^{2}(a-b)}{\left|a\right|+\left|b\right|}\leq1[/itex] no matter what the value of δ .

    But you need to find a δ so that [itex]\displaystyle \frac{\sin^{2}(a-b)}{\left|a\right|+\left|b\right|}\leq\varepsilon[/itex], whenever [itex]\displaystyle 0<\sqrt{a^2+b^2}<\delta\,.[/itex]
     
  7. Jun 26, 2012 #6
    I'm not sure how I can go about that. Where should I start looking for a delta ?
     
  8. Jun 26, 2012 #7

    SammyS

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    Well, you know that [itex]\displaystyle \left|\sin(a-b)\right|\leq\left|a \right|+\left|b\right|\,,[/itex]

    so [itex]\displaystyle \frac{\sin^{2}(a-b)}{\left| a\right|+\left|b\right|}\leq\left|a\right|+\left|b\right|\ .[/itex]

    And as is usually done, start with ε and \work your way to δ.

    Then for the proof you write-up, reverse the order.

    Typo fixed in Edit.
     
    Last edited: Jun 26, 2012
  9. Jun 26, 2012 #8
    Okay, so this is where I am. [itex]0\leq\sqrt{a^{2}+b^{2}}^{2}\leq\delta^{2}[/itex]
    So
    [itex]0\leq\left|a^{2}+b^{2}\right|\leq\delta^{2}[/itex]
    and because
    [itex]\left|a^{2}+b^{2}\right|\leq \left|a^{2}\right|+\left|b^{2}\right|[/itex]
    Then
    [itex]0\leq\left|a^{2}\right|+\left|b^{2}\right|\leq\delta^{2}[/itex]
    and I would somehow slide in that
    [itex]\frac{\sin^{2}(a-b)}{\left|a\right|+\left|b\right|}[/itex]≤[itex]\left|a\right|+\left|b\right|[/itex]
    I feel like I could use the triangle inequality: [itex]\left|a\right|+\left|b\right|\geq \sqrt{a^{2}+b^{2}}[/itex]
    but I am I allowed to squeeze it in between the [itex]\sqrt{a^{2}+b^{2}}[/itex]and δ?
    because I've so far only squeezed stuff in between [itex]\sqrt{a^{2}+b^{2}}[/itex] and 0
     
  10. Jun 26, 2012 #9

    SammyS

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    Can you show that [itex]\displaystyle 2\sqrt{a^2+b^2}\ge\left|a\right|+\left|b\right|\ ?[/itex]
     
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