# 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.

• Try $\tan^{-1}(z)=z-\frac{z^3}{3}+\frac{z^5}{5}-\frac{z^7}{7}.........$ Commented Feb 13, 2016 at 14:24
• Thank you but the Taylor's theorem for more than one variable isn't in the unit of these exercises yet, so, we aren't supposed to use it in spite of the resolution that way. Commented Feb 13, 2016 at 14:40

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 $0$.

• +1 for both the use of "sought" and the natural use of polar coordinates. Very well-(s|th)ought. Commented Feb 13, 2016 at 14:34
• @ClementC. Thank you :) Commented Feb 13, 2016 at 14:50
• Very nice solution. Commented Feb 13, 2016 at 14:58

$-\frac{x^2+y^2}{2}\leq xy\leq \frac{x^2+y^2}{2}$ 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: $$-\frac{r^2}{2r}\leq\frac{\arctan\frac{-r^2}{2}}{r}\leq \frac{\arctan{(xy)}}{\sqrt{x^2+y^2}}\leq \frac{\arctan\frac{r^2}{2}}{r}\leq\frac{r^2}{2r}$$

• Anyone cares to justify his downvote ? Commented Feb 13, 2016 at 14:38
• This is the best answer if you ask me. No need to use dodgy polar coordinates. Commented Feb 13, 2016 at 17:16

Use polar coordinates and L'Hospital's rule:

$$\lim\limits_{(x,y) \to (0,0)} \frac{\arctan(xy)}{\sqrt{x^2+y^2}} = \lim\limits_{r \to 0} \frac{\arctan(r^2 \cos(\theta) \sin(\theta))}{\sqrt{(r \cos(\theta))^2+(r \sin(\theta))^2}} = \lim\limits_{r \to 0} \frac{\arctan(r^2 \cos(\theta) \sin(\theta))}{r} = \lim\limits_{r \to 0} \frac{2r \cos(\theta) \sin(\theta))}{r^2 \sin^2(\theta)\cos^2(\theta) + 1} = 0$$

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 $0.$

• It's not tan(x·y) but arctan(x·y) Commented Feb 13, 2016 at 18:19
• Thank you. I've edited to correct it.
– zhw.
Commented Feb 13, 2016 at 19:17

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 $0$.