Epsilon-Delta Proof of Limit in R2

[ChemEng1]

Homework Statement

Examine the limit behavior on (ℝ², 2-norm) for f(λ,μ)=\frac{λμ}{|λ|+|μ|} 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.

Homework 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.

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 {a_n}={(\frac{1}{n},\frac{1}{n})}. Observe (0,0) is not an element of {a_n}. Let a=(0,0). d(a_n,a)=\sqrt{\left(\frac{1}{n}-0\right)^{2}+\left(\frac{1}{n}-0\right)^{2}}\Rightarrow\frac{\sqrt{2}}{n}=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→\sqrt{\left(\frac{λμ}{|λ|+|μ|}-f(0,0)\right)^{2}}<ε.

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

How far off track am I on this?

 

[LCKurtz]

Homework Statement

Examine the limit behavior on (ℝ², 2-norm) for f(λ,μ)=\frac{λμ}{|λ|+|μ|} defined on ℝ²\{(0,0)} at (λ,μ)=(0,0).

I think the limit of f(λ,μ) at (0,0) is (0,0)

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