# Epsilon Delta proof of a 2variable limit. Is my proof valid?

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1. Jan 23, 2016

### AutumnWater

1. The problem statement, all variables and given/known data
Use the epsilon delta definition to show that lim(x,y) -> (0,0) (x*y^3)/(x^2 + 2y^2) = 0
2. Relevant equations
sqrt(x^2) = |x| <= sqrt(x^2+y^2) ==> |x|/sqrt(x^2+y^2) <= 1 ==> |x|/(x^2+2y^2)?

3. The attempt at a solution
This limit is true IFF for all values of epsilon > 0, there exists a delta such that:

0<sqrt(x^2+y^2)<delta ==> |(x*y^3)/(x^2+2y^2)| < epsilon

When checking my answers, I'm stuck at this step:
Can: |x|/sqrt(x^2+2y^2) <= 1 imply that |x|/(x^2+2y^2) will also be <= 1?
I'm skeptical as it seems wrong, but if this step holds, then the rest of my proof can be as follows:

(|x|/(x^2+2y^2))*|y^3| <= |y^3| = |(y^2)^(3/2)| <= (x^2 + y^2 ) ^ (3/2) < (delta)^3

thus

let delta = (epsilon)^(1/3) and we get:

sqrt(x^2+y^2)<epsilon^(1/3) ==> (x^2+y^2)^(3/2) < epsilon ==> (y^2)^(3/2) = y^3 < (x^2+y^2)^(3/2) < epsilon;

Since |x|/(x^2+2y^2) <= 1, (|x|/(x^2+2y^2)) * |y^3| < |y^3| < epsilon;

and we have f(x) < epsilon.

2. Jan 23, 2016

### Samy_A

Can: |x|/sqrt(x^2+2y^2) <= 1 imply that |x|/(x^2+2y^2) will also be <= 1?
No. (Take x=0.5 and y=0.1 for example)

Hint: $\frac{1}{x²+2y²}\leq \frac{1}{2y²}$

3. Jan 23, 2016

### AutumnWater

thanks, I'll go review my polar coordinates chapters first :P

4. Jan 23, 2016

### Samy_A

That works too.

5. Jan 23, 2016

### AutumnWater

so we have: f(x,y) = (xy^3)/(x^2+2y^2)
since 1/(x^2+2y^2) <= 1/2y^2
f(x,y) <= (xy^3)/2y^2 where (xy^3)/2y^2 = xy/2 = p^2 sin(2theta) / 4 (let that = g(p,theta))
since sin(2theta) is between [-1, 1],
-p^2/4 <= g(p,theta) <= p^2/4
and f(x,y) is <= g(p,theta)
does that mean I should set delta = 2sqrt(epsilon) here then?

We need 0<p<delta to imply g(p,theta)< epsilon, since we know g(p,theta)< p^2/4, if we set delta = 2sqrt(epsilon), 0<p<2sqrt(epsilon) ==> (p^2)/4 < epsilon, then it works?

So in the end we have:
1) 0<sqrt(x^2+y^2)<delta where sqrt(x^2+y^2) = p, set delta = 2sqrt(epsilon)

2) |f(x)| <= g(p,theta) <= p^2/4 < epsilon

since p<2sqrt(epsilon) ==> (p^2/4) < epsilon

we have a connection between 1) and 2) here.

Last edited: Jan 23, 2016
6. Jan 24, 2016

### SammyS

Staff Emeritus
I assume you mean 1/(2y2) on the right hand side.

1/(x^2+2y^2) ≤ 1/(2y2) implies that x2 ≥ y2. Did you really mean that?

If you're going to change this to polar coordinates, Nothing is more convenient to convert than x2 + y2 .

After all, x2 + y2 = r2 .

Somehow I missed seeing the coefficient of 2 on the y2 in the denominator, so ignore this, which it seems you have. (thankfully!)

Last edited: Jan 24, 2016
7. Jan 24, 2016

### AutumnWater

yes I meant 1/(x^2+2y^2) <= 1/(2y^2) implies (xy^3)/(x^2+2y^2) <= (xy^3)/(2y^2)

Last edited: Jan 24, 2016
8. Jan 24, 2016

### pasmith

You need some absolute value signs: the inequality is reversed when $xy < 0$.

9. Jan 24, 2016

### AutumnWater

Ok thanks.

Here I got 2 different deltas, would it be safe to say both are correct, only one is even more closer than the other one?
If looking at the initial equation directly:
We have:
$$|(xy^3)/(x^2+2y^2)| = [(p^2*(sin(theta))^2*sin(2theta))/2(1+(sin(theta))^2)] <= (p^2)/2 < epsilon$$
so we set delta = sqrt(2epsilon)

If looking at the equations from the perspective of $$|xy^3|/2y^2$$: then delta's $$2sqrt(epsilon)$$ according to previous post.

10. Jan 26, 2016

### haruspex

You could make it a lot simpler by substituting $y=z/\sqrt 2$ before going polar.

11. Jan 26, 2016

### AutumnWater

would it? whilst there will be a coefficient of $$1/(2^{3/2})$$ on the numerator of the f(x) only when doing the z substitution, the equation for delta wouldn't be p=sqrt(x^2 + y^2) anymore, instead it would turn out to be sqrt(x^2 + (z^2)/2) wouldn't it?

so $$|(x^{2} + (z^{2})/2))^{1/2}| <= delta$$ needs to imply $$|((xz^{3})/(2^{3/2}) / (x^{2} + z^{2})| < epsilon$$ ?

in polar form that would amount to:

sqrt(5p^2/4) < delta needs to imply that p^2/2 < epsilon

(sorry, I will get a hang of latex codes soon)

Last edited: Jan 26, 2016
12. Jan 26, 2016

### haruspex

So make delta a bit smaller for the given epsilon.

13. Jan 26, 2016

### AutumnWater

oops, the p^2/2 < epsilon earlier didn't include the $$1/(2^{3/2})$$

so if it was $$(1/(2^{3/2})) * (p^2/2) < epsilon ==> p^2/2 < 2sqrt(2)epsilon ==> p < 2 (2^{1/4})\sqrt(epsilon)$$

and $$\sqrt(5)p/2 < delta ==> p < 2(delta)/\sqrt(5)$$

make ps equal, and I got $$\sqrt(5)(2^{1/4})\sqrt(epsilon) = delta$$ which is about 2.659 times $$\sqrt(epsilon)$$, bigger than the other two...

so far I've had: $$\sqrt(2epsilon)$$, $$2\sqrt(epsilon)$$, and now $$(2^{1/4})\sqrt(5epsilon)$$