# Multivariable limit $\lim\limits_{(x,y)\to(0,0)}\frac{x^2y^2}{|x|^3+|y|^3}=0$

I need to prove

$$\lim\limits_{(x,y)\to(0,0)}\frac{x^2y^2}{|x|^3+|y|^3}=0$$

I sort of know how to do it using polar coordinates, but I was trying to find an upper bound. Any ideas? I also wouldn't mind if someone could give me a hand formalizing the proof using polar coordinates. Thanks!

You could apply the AM-GM inequality to the denominator.

• Could you show me please how that would help? – Lessa121 Mar 24 '14 at 14:43
• @LunaSage It would give you $|x|^3+|y|^3 \ge 2|x|^{3/2}|y|^{3/2}$. Since the numerator is equal to $|x|^2|y|^2$ we can therefore bound the fraction by $(|x||y|)^{1/2}/2$ which goes to zero. – Jyotirmoy Bhattacharya Mar 24 '14 at 15:03

Hint: is enough to prove that $\cos^3\theta+\sin^3\theta\ge$ const $>0$ for $\theta\in[0,\pi/2]$.

• This would be to show that the limit is independent of the value of theta? – Lessa121 Mar 24 '14 at 14:38
• Yes. You can bound the fraction easily. – Martín-Blas Pérez Pinilla Mar 24 '14 at 14:41
• There is one theorical thing I don't understand though. If I replace $x=r*cos(\theta)$ and $x=r*sin(\theta)$, so that when $(x,y)\to(0,0)$, then $r\to0$ isn't theta a constant? or theta can be anything? I mean, why do I have to prove that it should be independent of theta? – Lessa121 Mar 24 '14 at 14:42
• Independent of theta. – Martín-Blas Pérez Pinilla Mar 24 '14 at 14:42

Hint: Since $\frac{x}{1+x^3}\le\frac{2^{2/3}}{3}$ for $x\ge0$, we have \begin{align} \left|\frac{x^2y^2}{|x|^3+|y|^3}\right| &=\color{#00A000}{\frac{|y/x|}{1+|y/x|^3}}|y|\\ &\le\color{#00A000}{\frac{2^{2/3}}{3}}|y| \end{align}

Polar Coordinates

Show that $$\sin^2(\theta)\cos^2(\theta)=\frac14\sin^2(2\theta)\le\frac14$$ and $$\left|\sin^3(\theta)\right|+\left|\cos^3(\theta)\right|\ge\frac1{\sqrt2}$$ Then, since $x=r\cos(\theta)$ and $y=r\sin(\theta)$, we have \begin{align} \left|\frac{x^2y^2}{|x|^3+|y|^3}\right| &=\frac{r^4\left|\,\sin^2(\theta)\cos^2(\theta)\,\right|}{r^3(\left|\sin^3(\theta)\right|+\left|\cos^3(\theta)\right|)}\\ &\le\frac{1/4}{1/\sqrt2}r\\ &=\frac{\sqrt2}{4}r \end{align}