1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Limits with x and y

  1. Nov 17, 2009 #1
    1. The problem statement, all variables and given/known data
    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
    [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.

    [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
    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook