What is $\lim_{(x,y) \to (0,0)} \arctan(xy)/\sqrt{x^2+y^2}$? The limit is this: $$\lim\limits_{(x,y) \to (0,0)} \frac{\arctan(xy)}{\sqrt{x^2+y^2}}$$
It's not necessary to give a whole solution, I want the path to see how to solve it. 
I tried both with sequences characterization and definition of limit but I don't know many things to do with $\arctan(xy)$. I only know it's bounded $-{\pi \over 2} < \arctan(xy) < {\pi \over 2}$, it's odd and strictly increasing.
 A: Observe that you have
$$
\left|\arctan u\right|\leq|u|, \quad  |u|\leq1,
$$ then, switching to polar coordinates with $r=\sqrt{x^2+y^2}$, as $r\to 0$, you get

$$
\left|\frac{\arctan(xy)}{\sqrt{x^2+y^2}}\right|=\frac{\left|\arctan(r^2 \sin \theta \cos \theta)\right|}{r}\leq \frac{\left|r^2 \sin \theta \cos \theta\right|}{r}\leq r.
$$ 

The sought limit is equal to $0$.
A: $-\frac{x^2+y^2}{2}\leq xy\leq \frac{x^2+y^2}{2}$ and $\arctan$ is a monotonic increasing function, so $$\frac{\arctan\frac{-r^2}{2}}{r}\leq \frac{\arctan{(xy)}}{\sqrt{x^2+y^2}}\leq \frac{\arctan\frac{r^2}{2}}{r}$$ where $r=\sqrt{x^2+y^2}$.
Now, using L'Hospital's rule you get that both limits in the sandwich are $0$.
Or, you can use the inequality Oliver Oloa gave to get:
$$-\frac{r^2}{2r}\leq\frac{\arctan\frac{-r^2}{2}}{r}\leq \frac{\arctan{(xy)}}{\sqrt{x^2+y^2}}\leq \frac{\arctan\frac{r^2}{2}}{r}\leq\frac{r^2}{2r}$$
A: Use polar coordinates and L'Hospital's rule:
$$\lim\limits_{(x,y) \to (0,0)} \frac{\arctan(xy)}{\sqrt{x^2+y^2}} = \lim\limits_{r \to 0} \frac{\arctan(r^2 \cos(\theta) \sin(\theta))}{\sqrt{(r \cos(\theta))^2+(r \sin(\theta))^2}} = \lim\limits_{r \to 0} \frac{\arctan(r^2 \cos(\theta) \sin(\theta))}{r} = \lim\limits_{r \to 0} \frac{2r \cos(\theta) \sin(\theta))}{r^2 \sin^2(\theta)\cos^2(\theta) + 1} = 0$$
A: The absolute value of the expression is
$$\left |\frac{\arctan (xy)}{\sqrt {x^2+y^2}}\right| = \left |\frac{\arctan (xy)}{xy}\right|\cdot |y|\cdot \left |\frac{x}{\sqrt {x^2+y^2}}\right |\le \left |\frac{\arctan (xy)}{xy}\right|\cdot |y|\cdot 1.$$
Because $\lim_{u\to 0}(\arctan u)/u =1$ and $|y|\to 0,$ the desired limit is $0.$
A: Recalling from elementary geometry that the sine function satisfies the inequalities
$$x\cos(x)\le \sin(x)\le x$$
for $0\le x\le \pi/2$, then we can assert that
$$|\arctan(x)|\le |x|$$
for all $x$.  Therefore, we have
$$\left|\frac{\arctan(xy)}{\sqrt{x^2+y^2}}\right|\le \frac{|xy|}{\sqrt{x^2+y^2}}\le \sqrt{|xy|/2}$$
from which the coveted limit is seen to be $0$.
