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

Related problems: I, II. Here is how you advance

$$ \Bigg| \frac{xy}{\sqrt{x^2+y^2}}-0 \Bigg |\leq \frac{|x||y|}{\sqrt{x^2+y^2}} \leq \frac{ \sqrt{x^2+y^2} \sqrt{x^2+y^2} }{\sqrt{x^2+y^2}}=\sqrt{x^2+y^2}< \epsilon =\delta .$$

Note:

$$ |x| \leq \sqrt{x^2+y^2},\quad |y| \leq \sqrt{x^2+y^2}. $$

Convert into polar coordinates: so $x=r\cos\theta\space$ and $y=r\sin\theta.$

$\displaystyle\lim_{(x,y)\to(0,0)} \dfrac{xy}{\sqrt{x^2+y^2}}\\=\displaystyle\lim_{r\to 0}\dfrac{r\cos\theta\cdot r\sin\theta}{\sqrt{r^2\cos\theta+r^2\sin\theta}}\\=\displaystyle\lim_{r\to 0}\dfrac{r^2\sin\theta\cos\theta}{\sqrt{r^2(\sin^2\theta+\cos^2\theta)}} \text{Recall the Pythagorean Identity,}\sin^2\theta+\cos^2\theta=1.\\=\displaystyle\lim_{r\to 0}\dfrac{r^2\sin\theta\cos\theta}{\sqrt{r^2}}\\=\displaystyle\lim_{r\to 0}\space r\sin\theta\cos\theta\\=0$

You could do the following: $x^2+y^2\geq 2|xy|$, so $\sqrt{x^2+y^2}\geq \sqrt{2|xy|}$, therefore $$\left|\frac{xy}{\sqrt{x^2+y^2}}\right|\leq\frac{|xy|}{\sqrt{2|xy|}}=\frac{\sqrt{|xy|}}{\sqrt{2}},$$ which converges to $0$ as $x,y\to 0$.