Limit of two variables, proposed solution check: $\lim_{(x,y)\to(0,0)}\frac{xy}{\sqrt{x^2+y^2}}$ Does this solution make sense,
The limit in question:
$$
\lim_{(x,y)\to(0,0)}\frac{xy}{\sqrt{x^2+y^2}}
$$
My solution is this:
Suppose, $$ \sqrt{x^2+y^2} < \delta $$ therefore $$xy<\delta^2$$ So by the Squeeze Theorem the limit exists since $$\frac{xy}{\sqrt{x^2+y^2}}<\frac{\delta^2}{\delta}=\delta$$
Is this sufficient?
 A: Consider the continuous curve $C_{\alpha}(t)=(t,\alpha t)\subset\mathbb R^2$. Obviously $\lim C_{\alpha}(t)=(0,0)$ when $t\to 0$ and $$\lim_{t\to 0} f(C_{\alpha}(t))=0$$  so the limit of $f$ is probably exists. Now note that if $$||(x,y)-(0,0)||<\delta$$ then we have equivalently $$\sqrt{x^2+y^2}<\delta$$ and so $$|x|<\delta,~~|y|<\delta$$ Therefore if $z=\text{max}\{|x|,|y|\}$ so $z<\delta$ and $$\big|\frac{xy}{\sqrt{x^2+y^2}}-0\big|=\frac{|x||y|}{\sqrt{x^2+y^2}}\le \frac{z^2}{\sqrt{z^2+0}}=z<\delta$$
A: Here's a more direct solution.
We know $|x|,|y|\le\sqrt{x^2+y^2}$, so if $\sqrt{x^2+y^2}<\delta$, then
$$\left|\frac{xy}{\sqrt{x^2+y^2}}\right|\le\frac{\big(\sqrt{x^2+y^2}\big)^2}{\sqrt{x^2+y^2}}=\sqrt{x^2+y^2}<\delta.$$
A: I'm not sure I really follow your solution well - it doesn't seem to make sense to me (edit: see Cameron Buie's comment). Why not use polar coordinates by making the substitution $x = r\cos \theta, y = r \sin \theta$ though?
A: Another direct approach is to note that the square of the reciprocal of your function is $\frac{1}{x^2}+\frac{1}{y^2}$.
