Find $\lim_{(x,y)\to (0,0)} \frac{xy^2}{|x|+|y|}$

Question

Find $$ \lim_{(x,y) \to (0,0)} \frac{xy^2}{|x| + |y|}. $$

Attempt:

Putting $x = r\cos\theta,\; y = r\sin\theta$, $$ \lim_{(r, \theta) \to (0,0)} \frac{r^2 \sin^2\theta \cos\theta}{|\cos\theta| + |\sin\theta|} \leq \lim_{(r, \theta) \to (0,0)} \frac{r^2}{|\cos\theta| + |\sin\theta|} = 0 \qquad (?) $$

Please solve this question. Is the above approach correct?

Edit:

$$ \lim_{r\to 0} \frac{r^2 \sin^2\theta \cos\theta}{|\cos\theta| + |\sin\theta|} \leq \lim_{r\to 0} \frac{r^2}{|\cos\theta| + |\sin\theta|} = 0 $$

Answer 1

Check this:

$\left|\cfrac{xy^2}{|x|+|y|}\right|=|y|\cdot\cfrac{|x||y|}{|x|+|y|}\leq |y|\cdot\cfrac{(|x|+|y|)^2}{4(|x|+|y|)}=|y|\cdot\cfrac{|x|+|y|}{4}$

Which clearly goes to zero. In the second step we used the AM-GM inequality:

$ab\leq\left(\cfrac{a+b}{2}\right)^2$

for $a=|x|$, $b=|y|$

Answer 2

I would avoid polar coordinates altogether. Notice $$ \left|\frac{xy^2}{|x| + |y|}\right| = \frac{|x|}{|x|+|y|} y^2 \leq y^2 $$ So by the squeeze theorem, $\lim_{(x,y) \to (0,0)} \frac{xy^2}{|x| + |y|} = 0$.

Answer 3

For each $\theta\in[0,2\pi]$, $|\cos\theta|+|\sin\theta|\geqslant1$. Therefore $\dfrac{xy^2}{|x|+|y|}\leqslant r^2$, and so your limit is $0$ indeed. But it is not correct to assume that $(x,y)\to(0,0)$ is the same thing as $(r,\theta)\to(0,0)$.