Question

Can someone help me with htis epsilon delta proof please? My work in comments:

Answer 1

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.

Answer 2

hint:
if we rewrite such function, with polar coordinate, we get:
\[\Large f\left( {\rho ,\theta } \right) = {\rho ^2}\cos \theta {\left( {\sin \theta } \right)^3}\]

Answer 3

now, we can apply the epsilon-delta method for \(\rho\) variable, being \(\theta\) an established value

Answer 4

Ok thanks, I gotta go review my polar function chapters then.

Answer 5

sorry, I have misunderstood the function, please ignore my reply above

Answer 6

here is the right formula:
\[\Large f\left( {\rho ,\theta } \right) = \frac{{{\rho ^2}\cos \theta {{\left( {\sin \theta } \right)}^3}}}{{1 + {{\left( {\sin \theta } \right)}^2}}}\]

Answer 7

now, for fixed \(\theta\), \(f\) is a function of \(\rho\) only

Answer 8

so, when \(x,y \to 0\), then \(\rho \to 0\), being \(\rho= \sqrt {(x^2+y^2)}\)

Answer 9

Sorry, I haven't reviewed polar coordinates yet and can't follow this. but there's another hint I got from another forum: use (1/(x^2+2y^2)) <= 1/(2y^2) he said

Answer 10

yes! In that case we have to show that the new function \(g(x,y)\) checks this condition:
\[\Large \mathop {\lim }\limits_{x,y \to 0} g\left( {x,y} \right) = 0\]

Answer 11

where:
\[\Large g\left( {x,y} \right) = \frac{{x{y^3}}}{{2{y^2}}} = \frac{{xy}}{2}\]

Answer 12

now, using the polar coordinates, we have:
\[\Large g\left( {\rho ,\theta } \right) = {\rho ^2}\frac{{\sin \left( {2\theta } \right)}}{4}\]

Answer 13

so, we can write this:
\[\Large 0 \leqslant g\left( {\rho ,\theta } \right) \leqslant \frac{{{\rho ^2}}}{4}\]
and for \(\rho \to 0\), we get \(g \to 0\)

Answer 14

sweet, i'll review this question once i've done my reading, so it probably will take some time before I can understand and absorb all this, but thanks alot anyway!

Answer 15

:)

Answer 16

oops.. we can write this:
\[ \Large - \frac{{{\rho ^2}}}{4} \leqslant g\left( {\rho ,\theta } \right) \leqslant \frac{{{\rho ^2}}}{4}\]
and when \(\rho \to 0\) then \(g \to 0\)

Answer 17

since:
\[\Large - 1 \leqslant \sin \left( {2\theta } \right) \leqslant 1\]

Answer 18

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?

Answer 19

if we consider the function \(g\) then you are right, it is suffice to take \(\delta = \sqrt 2 \epsilon\), furthermore, we have to keep in mind that we are using the euclidean metric, so the neighbour about a point x is given by a circle centered at x

Answer 20

but since $$1/(x^2+2y^2) <= 1/2y^2$$ therefore $$xy^3/(x^2+2y^2) <= g(p,theta)$$, would it be ok to then use the g function's delta for f?

Answer 21

I'm thinking...

Answer 22

Ok I'll be right back after dinner thanks!

Answer 23

we can write this:
\[\large \begin{gathered}
f\left( {\rho ,\theta } \right) = \frac{{{\rho ^2}\cos \theta {{\left( {\sin \theta } \right)}^3}}}{{1 + {{\left( {\sin \theta } \right)}^2}}} = \frac{{{\rho ^2}{{\left( {\sin \theta } \right)}^2}}}{{1 + {{\left( {\sin \theta } \right)}^2}}}\frac{{\sin \left( {2\theta } \right)}}{2} \hfill \\
\hfill \\
- \frac{{{\rho ^2}}}{2} \leqslant \frac{{{\rho ^2}{{\left( {\sin \theta } \right)}^2}}}{{1 + {{\left( {\sin \theta } \right)}^2}}}\frac{1}{2} \leqslant f\left( {\rho ,\theta } \right) \leqslant \frac{{{\rho ^2}{{\left( {\sin \theta } \right)}^2}}}{{1 + {{\left( {\sin \theta } \right)}^2}}}\frac{1}{2} \leqslant \frac{{{\rho ^2}}}{2} \hfill \\
\end{gathered} \]

since:
\[\large - \frac{1}{2} \leqslant \frac{{\sin \left( {2\theta } \right)}}{2} \leqslant \frac{1}{2}\]

Answer 24

oops.. I have made a typo, here is the right inequality:
\[ \large - \frac{{{\rho ^2}}}{2} \leqslant - \frac{{{\rho ^2}{{\left( {\sin \theta } \right)}^2}}}{{1 + {{\left( {\sin \theta } \right)}^2}}}\frac{1}{2} \leqslant f\left( {\rho ,\theta } \right) \leqslant \frac{{{\rho ^2}{{\left( {\sin \theta } \right)}^2}}}{{1 + {{\left( {\sin \theta } \right)}^2}}}\frac{1}{2} \leqslant \frac{{{\rho ^2}}}{2}\]

Answer 25

so, it is suffice to choose \(\delta= \sqrt 2 \; \epsilon\), and the relation of limit, is proven

Answer 26

sqrt(2epsilon) or sqrt(2)*epsilon?

Answer 27

shouldn't we make p^2/2 < epsilon?

Answer 28

it is: \((\sqrt 2) \epsilon\)

Answer 29

yes! We have to solve this equation:
\[\huge \frac{{{\delta ^2}}}{2} = \epsilon \]

Answer 30

sorry:
\[\huge \delta = \sqrt {2\epsilon } \]

Answer 31

cool, thanks alot!

Answer 32

:)

Answer 33

Hi again, just unsure about the part I msged you about

how did -p^2 / 2 <= -p^2sin^2(theta)/(1+sin^2(theta)) appear?

Answer 34

step by step, I have used this identities:
\[\large \begin{gathered}
\cos \theta {\left( {\sin \theta } \right)^3} = {\left( {\sin \theta } \right)^2}\sin \theta \cos \theta = {\left( {\sin \theta } \right)^2}\frac{{2\sin \theta \cos \theta }}{2} = \hfill \\
\hfill \\
= {\left( {\sin \theta } \right)^2}\frac{{\sin \left( {2\theta } \right)}}{2} \hfill \\
\end{gathered} \]

Answer 35

oops.. these* identities

Answer 36

i understand the sin(2theta)/2 part, but how are the rest ?

Answer 37

it is the numerator

Answer 38

(sin(theta))^2 / (1+sin(theta)^2), how did that become a 1?

Answer 39

here are more steps:
\[\large \frac{{{\rho ^2}{{\left( {\sin \theta } \right)}^2}}}{{1 + {{\left( {\sin \theta } \right)}^2}}}\frac{{\sin \left( {2\theta } \right)}}{2} \leqslant \frac{{{\rho ^2}{{\left( {\sin \theta } \right)}^2}}}{{{{\left( {\sin \theta } \right)}^2}}}\frac{{\sin \left( {2\theta } \right)}}{2} = {\rho ^2}\frac{{\sin \left( {2\theta } \right)}}{2}\]

Answer 40

oh gotcha, thanks, so it's not like sin(theta)^2 / (1+sin(theta)^2) can equal 1 or anything, it's just smaller or equal to.

Answer 41

yes!