# Limit of $\frac{xy^3}{1+y^3}$ when $(x,y) \to (0,0)$

Show that

$$\lim_{(x,y)\to (0,0)} \frac{xy^3}{1+y^3} = 0$$

The only way I know of doing it is using squeeze theorem, but I couldn't find any function. Any hint?

Thanks!

• I don't see the problem, letting either x or y tend to zero will yield 0 as your limit... – Asier Calbet Sep 28 '14 at 19:14
• @Assaultous2 You don't get to treat $x$ and $y$ separately here. You have to consider approaching $(0,0)$ along all paths. – James Sep 28 '14 at 19:15
• Indeed, thanks guys! – Giiovanna Sep 28 '14 at 19:17
• @James I up voted Assaultous2's comment perhaps because I misunderstood him. What I first I understood from his comment is that the function is defined on $(0,0)$, so it's enough (depending on what the OP knows) to just replace $x$ and $y$ with $0$. – Git Gud Sep 28 '14 at 19:17
• I suppose that reading is also valid. I was working on the assumption that the OP doesn't get to appeal to continuity. – James Sep 28 '14 at 19:21

For every $|y|\leqslant\frac12$, $|1+y^3|\geqslant1-\left(\frac12\right)^3\geqslant\frac12$ hence $$\left|\frac{xy^3}{1+y^3}\right|\leqslant2\cdot|x|\cdot|y|^3\leqslant|x|\cdot|y|,$$ which should be enough to conclude.
You can use polar coordinates as $$x=r\cos\theta \ \ y=r\sin\theta$$ and the limit as $r\to 0$. The limit will be $$\lim_{(x,y)\to (0,0)} \frac{xy^3}{1+y^3} = \lim_{r\to 0} \frac{r^4\cos\theta\sin^3\theta}{1+r^3\sin^3\theta} = \frac 01=0$$