(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

This is my first attempt at a multivariate ε-δ proof. I don't know if I'm doing it right or wrong so I'm hoping someone can shed some light.

Prove :

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

2. Relevant equations

[itex]\forall[/itex]ε>0, [itex]\exists[/itex]δ>0 | 0<|x-1||y-1|<δ [itex]\Rightarrow[/itex] |x^2+y^2-2|<ε

3. The attempt at a solution

So we know : |x-1|<δ and |y-1|<δ

|x^2+y^2-2| = |x^2-1+y^2-1| ≤ |x^2-1|+|y^2-1| = |x+1||x-1| + |y+1||y-1| < δ|x+1| + δ|y+1|

Now let us express |x+1| and |y+1| in terms of δ so we can further simplify to our desired value of δ.

We can observe that : |x+1| < δ+2 and |y+1| < δ+2

So we continue forth : δ|x+1| + δ|y+1| < δ(δ+2) + δ(δ+2)

Now for convenience let us agree to allow δ to be bounded, say δ≤1 for simplicity.

Now we get : δ(δ+2) + δ(δ+2) ≤ 6δ ≤ ε

Therefore δ=min{1,ε/6}

So now we verify δ satisfies the definition.

[itex]\forall[/itex]ε>0, [itex]\exists[/itex]δ=min{1,ε/6}>0 | 0<|x-1||y-1|<δ [itex]\Rightarrow[/itex] |x^2+y^2-2|<ε

|x^2+y^2-2| = |x^2-1+y^2-1| ≤ |x^2-1|+|y^2-1| = |x+1||x-1| + |y+1||y-1| < δ|x+1| + δ|y+1|| < δ(δ+2) + δ(δ+2) ≤ 3ε/6 + 3ε/6 = ε

Thus δ=min{1,ε/6} [itex]\Rightarrow[/itex] |f(x,y)-L|<ε and the proof is concluded. I want to know if I'm doing this correctly or if I've missed something.

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Two dimensional limit proof

**Physics Forums | Science Articles, Homework Help, Discussion**