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Limits with x and y

  1. Nov 17, 2009 #1
    1. The problem statement, all variables and given/known data
    Compute:
    1)[tex]lim_{(x,y)\rightarrow (0,0)} \frac{x^3-y^3}{x^2+y^2}[/tex]
    2)[tex]lim_{(x,y)\rightarrow (0,0)} \frac{sin(xy)}{y}[/tex]

    3. The attempt at a solution
    1)
    [tex]lim_{(x,y)\rightarrow (0,0)} \frac{x^3-y^3}{x^2+y^2}[/tex]
    Not sure what to do here. Either the limit doesn't exist or it equals zero. I think it equals zero because in the special case where x=y we get 0/2x²=0. Also when x=0, we get the limit of -y where y approaches zero, which is zero. Same goes for y=0. But I don't know how to prove that the limit is zero in general.

    2)
    [tex]lim_{(x,y)\rightarrow (0,0)} \frac{sin(xy)}{y}=lim_{z\rightarrow 0} \frac{sin(z)}{\frac{z}{x}}=lim_{z\rightarrow 0} x\frac{sin(z)}{z}=lim_{z\rightarrow 0}x lim_{z\rightarrow 0} \frac{sin(z)}{z}=0*x=0
    [/tex]
    Are all steps correct? I'm not sure about the substitution and using the product rule for limits.
     
  2. jcsd
  3. Nov 17, 2009 #2
    you don't only consider the case when x and y go to zero "together", but you're basically on the right track.. there are 3 cases: 1) they both tend to zero at the same time, x=y. 2) x goes to zero first or 3) y goes to zero first. so you have axes x ,y and z and so the limit exists if there exists an epsilon > 0 then there also exists delta1 > 0 and delta2 > 0 with the property that if |x-a| < delta1 and |y-a| < delta2 then |f(x,y) - L| < epsilon .

    so, take the case where x=y and evaluate the limit as a "single variable". then you can "fix" x or y, then evaluate the limit as one approaches zero first then observe what happens when other goes to zero after
     
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