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Epsilon-Delta Proof of Limit in R2

  1. Apr 5, 2012 #1
    1. The problem statement, all variables and given/known data
    Examine the limit behavior on (ℝ2, 2-norm) for f(λ,μ)=[itex]\frac{λμ}{|λ|+|μ|}[/itex] defined on ℝ2\{(0,0)} at (λ,μ)=(0,0).

    I think the limit of f(λ,μ) at (0,0) is (0,0). I have tried multiple paths to (0,0) that seem to indicate this. (λ=μ, λ=μ2, et al.)

    However, I am lost trying to develop a rigorous ε, δ proof for this. Any help would be appreciated.

    2. Relevant equations
    Definition of Limit: Given metric spaces (X,d) and (Y,d') a mapping T:X→Y and a cluster point x0 of X, we say that T has a limit as x approaches x0 and that limit is y0 if given ε>0, there exists δ>0 such that d'(Tx,y0)<ε when 0<d(x,x0)<δ.

    3. The attempt at a solution
    Step 1. Show that (λ,μ)=(0,0) is a cluster point of ℝ2\{(0,0)}.
    Let A={ℝ2\{(0,0)}). Consider sequence {an}={([itex]\frac{1}{n},\frac{1}{n}[/itex])}. Observe (0,0) is not an element of {an}. Let a=(0,0). d(an,a)=[itex]\sqrt{\left(\frac{1}{n}-0\right)^{2}+\left(\frac{1}{n}-0\right)^{2}}\Rightarrow\frac{\sqrt{2}}{n}[/itex]=0 as n→∞. Therefore (0,0) is a cluster point of A.

    (Perhaps a density argument would be more consise way to argue this?)

    Step 2. Apply Definition of Limit.
    Pick an ε>0. Then for d'(f(λ,μ),y0)<ε→[itex]\sqrt{\left(\frac{λμ}{|λ|+|μ|}-f(0,0)\right)^{2}}[/itex]<ε.

    From this I do not know how to pull out the d((λ,μ),(0,0))<δ→[itex]\sqrt{(λ-0)^{2}+(μ-0)^{2}}[/itex]<δ to continue building this proof.

    How far off track am I on this?
     
    Last edited: Apr 5, 2012
  2. jcsd
  3. Apr 5, 2012 #2

    LCKurtz

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    You mean you think the limit is 0. Try squaring the fraction and work on that.
     
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