1. Not finding help here? Sign up for a free 30min 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!

Limit of z = f(x,y)

  1. Sep 28, 2013 #1
    1. The problem statement, all variables and given/known data
    use polar coordinates and L'hopital's rule to find the limit:
    Lim (x,y) -> (0,0) of (x2 + y2)*ln(x2+y2)





    3. The attempt at a solution
    I was told in class we couldn't use l'hopital because of the multivariable thing, and I was also told the coordinate switch from rectangular to polar wasn't possible. I'm not sure if the problem is a trick question, because if I can't use L'hopital I would say the limit doesn't exist at (0,0) but how am i supposed to do this before the ideas of partial derivatives are introduced? How am I supposed to use L'Hopital's rule to help that is.
    and when do I use the polar switch? before or after I take L'Hopital?
    when I do switch over to polar should (x,y) become (rcos(θ),rsin(θ)) -> (0,0) or do I have to switch over from (0,0) to some other angle? am I picking an angle that simultaneously makes cos and sin zero? I don't think that one is possible, so perhaps I switch to something like (pi/2, pi) or (0,2pi) how do I choose between those? I could see sign errors arising if I make the wrong choice.

    Thank you for your time.
     
  2. jcsd
  3. Sep 28, 2013 #2

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    "and I was also told the coordinate switch from rectangular to polar wasn't possible."
    Who told you such a thing??
    It is wrong; you can.
     
  4. Sep 28, 2013 #3
    my professor, he probably was referring to the problem at hand, and that it wouldn't help get the correct answer. Maybe he misunderstood what I was asking him. So if I do use polar coordinates
    lim (rcos(t), rsin(t)) -> (0,0) [rln(r)] = 0ln(0) still undefined, differentiate wrt r and I get ln(r) + 1 this seems to work since the answer is 1. I need r = 1 so ln(1)= 0 hmm
     
  5. Sep 28, 2013 #4
    First, notice that when you switched to ##(rcos \theta, rsin \theta)## you put yourself back in one variable, so you could use L'Hospital's Rule.

    Now your confusion about the angles. The ##\theta## is whatever is correct for the vector (x,y). For example what angle does the vector (1,1) make with the x-axis? That is your ##\theta## . The vector (2,2) makes the same angle, so what about it differs from (1,1) in polar coordinates? What is ##\theta## for the vector (2,1)?

    Now your function ##(x^2 + y^2)*log(x^2 + y^2)## becomes what in polar coordinates? It is the function which is going to 0, not the vector components.

    See if you can make some progress now.
     
  6. Sep 29, 2013 #5

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    I believe you must have misunderstood what your professor was saying.

    The problem is perfectly well solvable by using first coordinate transformation, and then L'Hopital's rule (for infinity/infinity).

    rln(r) may be written as ln(r)/(1/r), upon which you may use L'Hopital
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Limit of z = f(x,y)
  1. Limit f(x,y) (Replies: 8)

  2. F(x, y, z) ? (Replies: 2)

  3. Limit of f(x,y) (Replies: 10)

Loading...