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Limits (another error in the question?)

  1. Feb 24, 2007 #1
    1. The problem statement, all variables and given/known data

    a) Prove that;

    [tex]\lim_{(x,y)\rightarrow(0,0)} \frac{x^2y}{x^2+y^2} = 0[/tex]

    b) Prove that if [tex]\lim_{(x,y)\rightarrow(0,0)} f(x,y) = L_1[/tex] and [tex]\lim_{(x,y)\rightarrow(0,0)} f(x,y) = L_2[/tex], then [tex]L_1=L_2[/tex]

    c) Using the statement proven in 5b, prove that

    [tex]\lim_{(x,y)\rightarrow(0,0)} \frac{xy}{x^2+y^2}[/tex]

    Does NOT exist.

    2. The attempt at a solution


    [tex]f(0,y) = \frac{0}{y^2} = 0[/tex]

    [tex]f(x,0) = \frac{0}{x^2} = 0[/tex]

    From those 2 directions, the limit is the same, so;

    [tex]\lim_{(x,y)\rightarrow(0,0)} \frac{x^2y}{x^2+y^2} = 0[/tex]


    I have no idea how to do that, it seems to evident !


    Does NOT exist ? But...

    [tex]f(0,y) = \frac{0}{y^2} = 0[/tex]

    [tex]f(x,0) = \frac{0}{x^2} = 0[/tex]

    It's exactly the same thing as in a), the limit DOES exist and it is;

    [tex]\lim_{(x,y)\rightarrow(0,0)} \frac{xy}{x^2+y^2} = 0[/tex]
  2. jcsd
  3. Feb 24, 2007 #2

    Gib Z

    User Avatar
    Homework Helper

    Ahh for b), seeing as they are the same limit, they must evaluate to the same thing. L1=L2
  4. Feb 24, 2007 #3
    ops, I got it wrong. The limit in c doesn't exist because at (x,mx);

    [tex]\lim_{(x,y)\rightarrow(0,0)} \frac{mx^2}{x^2(1+m)} = \frac{m}{m+1}[/tex]

    However, this leads me to 2 questions;

    I just proved the limit in c doesn't exist, why on earth would I need, or even, how could I use the statement in b ?

    Also, is my "proof" that the limit in a equal 0 sufficient ? Even at (x,mx) the limit is still 0, but still, is this a proof that the limit is 0 ?
    Last edited: Feb 24, 2007
  5. Feb 25, 2007 #4
    The logic in problem c) is that since you have two different limits you contradicted b) so that the limit does not exist.

    b) is just saying that if the (some) limit exists, then it is unique. Part b) can be resolved by general limit arguments, not talking just about function of two variables x,y.
  6. Feb 25, 2007 #5
    You answer to a) does not account for (x,y) approaching (0,0) in any arbitrary path. What happens as the distance r from (0,0) goes to zero? (Hint: remember polar coords?)
  7. Feb 25, 2007 #6


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    Staff Emeritus
    Science Advisor

    I think (b) is badly stated. I would interpret what they say as "the limit is unique". That's easy to prove. Suppose there existed two different limits, L1, L2. Take [itex]\epsilon= |L_1- L_2|/2[/itex]. Show that no [itex]\delta[/itex] makes both [itex]|f(x)- L_1|< \epsilon[/itex] and [itex]|f(x)- L_2|< \epsilon[/itex] for (x,y) such that [itex]\sqrt{(x^2+ y^2}< \delta[/itex].

    However, that's clearly not what's meant. What they mean is that if the limit itself exists, then the limits along any path must be the same. Unfortunately, the way it is stated implies that the "limit" means the limit "along some path" which is not true.

    As gammamcc said, showing that the limit along two different paths gives the same thing does not prove that the limit exists. you must show that f(x,y) is close to some L as long as (x,y) is close to (0,0). Again, as gammamcc said, consider changing to polar coordinates. That way r itself measures the distance from (0,0). If the limit, as r goes to 0, does not depend upon [itex]\theta[/itex], then the limit exists.
  8. Feb 25, 2007 #7
    Great, thank you both, I have my answer.
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