Show that $$\lim_{(x,y) \to (0, 0)}\frac{xy^2}{x^2+y^2} = 0$$
I have tried switching to polar coordinates but I'm not getting a single term. This is what I did.
Putting $$x=r\sin θ,\quad y=r\cos θ$$ we obtain
$$\left|\frac{xy^2}{x^2+y^2}\right|=|r\cos^2 θ \sin θ| =|r\sin θ(1-\sin^2 θ)| = |r\sin θ-r\sin^3 θ|$$
Using the change of variables $$\begin{align*}x &= \rho\cos\theta\\ y &= \rho\sin\theta \end{align*}$$
we obtain $$\lim_{\rho \to 0} \rho\cos\theta\sin^2\theta = 0,$$ since $\cos\theta$ and $\sin^2\theta$ are bounded.
Let $\epsilon > 0$ then if $\delta = \epsilon$ we have whenever $|x|<\delta$ and $|y|<\delta$ that
${|xy^2| \over x^2 + y2} <= {|x|(x^2+y^2)\over x^2+y^2} = |x| < \delta = \epsilon$
$|x| \le \sqrt{x^2+y^2}$ and $|y| \le \sqrt{x^2+y^2}$. So what is the absolute value of your expression in question? In general don't try switching to polar coordinates if the expression will not simplify.
