Epsilon-delta definition for a multivariable limit I'd like to prove the following limit to be zero using the epsilon-delta definition:  $\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{|x|^3+|y|^3}$
I tried to estimate |$\frac{x^2y^2}{|x|^3+|y|^3}$| to be $\lt\epsilon$ and thus find the fitting $\delta$. But I couldn't figure out the steps I need to take. I think the norm $\Vert.\Vert_3$ can be used here somehow. I can get rid of the $x^2y^2$ term like so:
Let $\epsilon\gt0$ and assume $\Vert X \Vert \lt1$, so $|x|\lt1$ and $|y|\lt1$.
Then $|\frac{x^2y^2}{|x|^3+|y|^3}-0| \lt |\frac{1}{(|x|^3+|y|^3)}|=\frac{1}{\Vert X \Vert_3}$
I would need to get the norm to the numerator and use $\Vert X \Vert_3 \le \Vert X \Vert_2$ to find $\delta$ so that $|\frac{x^2y^2}{|x|^3+|y|^3}-0| \lt \epsilon$ when $\Vert X-0 \Vert_2 \lt \delta$.
 A: When one approach doesn't give you what you need, it is time to abandon it and try something else. 
Divide the neighborhood of $0$ into two parts: where $|x| \le |y|$ and where $|y| \le |x|$. In the first region, divide top and bottom by $|y|^3$. Then the denominator is trapped between 1 and 2, which allows you to find bounds on the entire fraction. In the other region, divide by $|x|^3$ to do the same. Use the combined bounds to prove your theorem.
A: Since all norms are equivalent on $R^2$, there exist positive constants $c_1,c_2$ such that:
$$||(x,y)||_3\leq ||(x,y)||_2 \leq c_1||(x,y)||_3$$
and also 
$$||(x,y)||_1\leq c_2||(x,y)||_2$$
where, as usual,
$$||(x,y)||_p = \left(|x|^p+|y|^p\right)^{1/p}$$
Let $\epsilon > 0$. Take $\delta={\epsilon\over (c_1c_2)^4}$. Assume $||(x,y)||_2<\delta$. Then
$$|x||y|\leq (|x|+|y|)^2 =||(x,y)||_1^2\leq c_2^2||(x,y)||_2^2\leq c_2^2c_1^2||(x,y)||_3^2 $$
Squaring both sides:
$$x^2y^2\leq (c_2c_1)^4||(x,y)||^4_3$$
whence, upon dividing by $||(x,y)||_3^3$,
$${x^2y^2\over |x|^3+|y|^3}\leq (c_1c_2)^4||(x,y)||_3\leq(c_1c_2)^4||(x,y)||_2<(c_1c_2)^4\delta=\epsilon$$
thus proving that the limit is zero.
