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Homework Help: Prove the limit exists for xy/(sqrt(x^2+y^2)

  1. Mar 20, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the limit, if it exists, or show that the limit does not exist.
    Stewarts Calculus 7th edition 14.2 Q13.

    2. Relevant equations

    3. The attempt at a solution


    The limit along any line through (0,0) is 0, as well as along other
    paths through (0,0) such as [itex]x=y^{2}[/itex] and [itex]y=x^{2}[/itex]

    Let [itex]\epsilon>0[/itex] We want to find [itex]\delta>0[/itex] such that

    if [itex]0<\sqrt{x^{2}+y^{2}}<\delta[/itex] then [itex]\left|\frac{xy}{\sqrt{x^{2}+y^{2}}}-0\right|<\epsilon[/itex]

    Edit: Still can't get the latex working
    Last edited: Mar 20, 2013
  2. jcsd
  3. Mar 20, 2013 #2


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    Science Advisor

    Change to polar coordinates. That way the distance from (0, 0) depends only on r. If the limit, as r goes to 0, is the same, for all [itex]\theta[/itex], then the limit exists and is that value.
  4. Mar 20, 2013 #3
    So then [tex]f(x,y)=\frac{r\cos\theta r\sin\theta}{r^{2}}=r\cos\theta\sin\theta[/tex]

    That's pretty cool! So I can just do this for most functions right?

    Unfortunately, I have to prove the above using the definition of limits (as that's what we covered) for exams.
    Last edited: Mar 20, 2013
  5. Mar 20, 2013 #4


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    Well, the definition says, given [itex]\epsilon> 0[/itex], there exist [itex]\delta>0[/itex] such that if the [itex]\sqrt{x^2+ y^2}= r< \delta[/itex] then [itex]|f(x, y)|< \delta[/itex]. In polar coordinates, as you show, that is [itex]|r cos(\theta)sin(\theta)|< \epsilon[/itex] and you know that [itex]cos(\theta)|\le 1[/itex] and [itex]sin(\theta)\le 1[/itex].
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