Show $\lim\limits_{(x,y)\to(0,0)} \frac{xy}{\sqrt{x^2+y^2}}=0$ using Delta Epsilon Proof

Question

I tried following this answer (https://math.stackexchange.com/a/542303/86425), but I cannot see how this shows the limit is 0. Can someone please provide guidance

Answer 1

Note that $\sqrt{x^2 + y^2} \geq \sqrt{y^2} = |y|$. Then $$ \left|\frac{xy}{\sqrt{x^2 + y^2}}\right| = |x|\frac{|y|}{\sqrt{x^2 + y^2}} \leq |x| $$ so $$ \lim_{(x,y) \to (0,0)}\left|\frac{xy}{\sqrt{x^2 + y^2}}\right| \leq \lim_{(x,y) \to (0,0)} |x| = 0 \, . $$

Personally, I prefer thinking of these problems in polar coordinates. So let $x = r \cos\theta$, $y = r \sin \theta$ where $r = \sqrt{x^2 + y^2}$. Then \begin{align*} \lim_{(x,y) \to (0,0)} \frac{xy}{\sqrt{x^2 + y^2}} &= \lim_{r \to 0} \frac{r^2 \cos \theta \sin \theta}{r} = \lim_{r \to 0} r \cos \theta \sin \theta = 0 \, . \end{align*}

Answer 2

Use the inequality:$xy\leq\frac{x^2+y^2}{2}$

Answer 3

$z=\frac{xy}{\sqrt{x^2+y^2}}\Longrightarrow 1/z^2=1/x^2+ 1/y^2\Longrightarrow\lim\limits_{(x,y)\to(0,0)} 1/z^2=\infty\Longrightarrow\lim\limits_{(x,y)\to(0,0)} z=0.$