Is My δ Correct for lim (x,y)→(0,0) f(x,y) = 0?

[STEMucator]

Homework Statement

Show that : (x^4+y^4)/(x^2+y^2) < ε if 0 < x^2 + y^2 < δ^2 for a suitably chosen δ depending on ε.

Homework Equations

$\forall$ε>0, $\exists$δ>0 | 0 < (x^2 + y^2)^(1/2) < δ $\Rightarrow$ |f(x,y) - L| < ε

Obviously here were dealing with lim (x,y)→(0,0) f(x,y) = 0 so the following statement is equivalent and more convenient to use in my opinion:

$\forall$ε>0, $\exists$δ>0 | 0 < |x|,|y| < δ $\Rightarrow$ |f(x,y) - L| < ε

The Attempt at a Solution

So we know : |x| < δ $\Rightarrow$ x^2<δ^2 and also |y|<δ $\Rightarrow$ y^2<δ^2

And using the triangle inequality we also consider : |x^4 + y^4| ≤ |x|^4 + |y|^4

So putting those together we observe :

|f(x,y) - L| = |(x^4+y^4)/(x^2+y^2)| ≤ (|x|^4 + |y|^4)/(|x|^2 + |y|^2) < 2δ^4/2δ^2 = δ^2 ≤ ε

$\Rightarrow$ δ = $\sqrt{ε}$

Now that I have my δ, I could go through and prove that it was the right δ, but I have one problem. The book says that δ = $\sqrt{ε/2}$ so I'm wondering where I went wrong or is this a typo in the book? If it helps I also tried using the other statement 0 < (x^2+y^2)^(1/2) < δ and got the right answer, but I'm not sure why I'm wrong about this other method?

Thanks.

 

[christoff]

Although I agree with your choice of $\delta=\sqrt{\epsilon}$, I disagree with some of your calculations.

Homework Equations

$\forall$ε>0, $\exists$δ>0 | 0 < (x^2 + y^2)^(1/2) < δ $\Rightarrow$ |f(x,y) - L| < ε

Obviously here were dealing with lim (x,y)→(0,0) f(x,y) = 0 so the following statement is equivalent and more convenient to use in my opinion:

$\forall$ε>0, $\exists$δ>0 | 0 < |x|,|y| < δ $\Rightarrow$ |f(x,y) - L| < ε

Your statement of limit is incorrect here. If should read $0< ||(x,y)||<\delta \Rightarrow |f(x,y)|<\epsilon.$

Where I have taken your L to be zero, as is the question.

The Attempt at a Solution

So we know : |x| < δ $\Rightarrow$ x^2<δ^2 and also |y|<δ $\Rightarrow$ y^2<δ^2

This is true, but it is not necessary that we require |x| < δ and |y| < δ since this follows whenever $||(x,y)||<\delta$. Again, this is just a clarification. I more so wish to emphasize that whatever limit definition you used above was incorrectly copied down.

And using the triangle inequality we also consider : |x^4 + y^4| ≤ |x|^4 + |y|^4

This is fine. However, triangle inequality is not required since x^4 and y^4 are both positive.

So putting those together we observe :

|f(x,y) - L| = |(x^4+y^4)/(x^2+y^2)| ≤ (|x|^4 + |y|^4)/(|x|^2 + |y|^2) < 2δ^4/2δ^2 = δ^2 ≤ ε

$\Rightarrow$ δ = $\sqrt{ε}$

The second to last inequality does not follow. In general, we have the following:
$$(x^2+y^2)<\delta \Rightarrow 1/\delta<1/(x^2+y^2).$$
However, you have (falsely) assumed
$$(x^2+y^2)<\delta \Rightarrow 1/(x^2+y^2)<1/\delta$$
or something of that form.

A quick way to prove the end result would be to, first, drop those abolute values (since |x|^2=x^2, etc.), and then notice that $x^4+y^4=(x^2+y^2)^2-2x^2y^2$, and work from there.