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How does one find the limit of

$$\lim\limits_{(x,y) \to (0,0)} \dfrac{|xy|}{\sqrt{x^2 + y^2}}$$?

Can someone justify the steps they make? The answers in my book involves using some smart inequality that I've never seen before and could only say it resembles the AM-GM inequality

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Is the denominator $s^2$ or $x^2$? Also, try converting to polar coordinates; not sure if it will work, but it may. – Daryl Feb 5 '13 at 22:22
It is $x^2$ sorry and yes! Polar coordinates seem to does the trick! Thanks – Hawk Feb 5 '13 at 22:28
I will post an answer below as well for completeness. – Daryl Feb 5 '13 at 22:29
Yeah thanks and I'll accept it afterwards. – Hawk Feb 5 '13 at 22:31
up vote 3 down vote accepted

Transforming to polar coordinates, $x=r\cos\theta$ and $y=r\sin\theta$, gives the limit $$\lim\limits_{r\rightarrow0^+}\frac{r^2|\sin(2\theta)|}{2r},$$ which is easily evaluated to be $0$.

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Does it really matter if $r \to 0^+$ or $-$? – Hawk Feb 5 '13 at 22:36
No. However mathematically, since $r=\sqrt{x^2+y^2}\geq 0$, technically $\lim\limits_{r\rightarrow0}f(r)$ doesn't exist as $r$ is not defined for negative numbers. Hence, I included explicitly that it is for positive $r\rightarrow0$. – Daryl Feb 5 '13 at 22:41
I thought by convention $r$ can be negative because $r^2 = x^2 + y^2$ – Hawk Feb 5 '13 at 22:43
$r$ can never be negative. $r = \sqrt{x^2+y^2}$ by definition, and hence by definition of $\sqrt{(\cdot)}$, it must be positive. $r^2 = x^2+y^2$ by consequence of the definition of $r$. – Emily Feb 5 '13 at 23:07
There are two conventions for graphing polar curves. They go as follows. (i) $r=-2$, $\theta=\pi/3$ is not allowed or (ii) the point $r=-2$, $\theta=\pi/3$ is the point obtained thus: graph $r=2$, $\theta=\pi/3$ as usual, then reflect the result in the origin, or equivalently rotate by a half-turn. So in a homework problem, one has to be aware of which convention is the one used in the course. In our case, we can choose either convention, and $r\ge 0$ is more convenient. – André Nicolas Feb 5 '13 at 23:08

Hint: Note that,

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

and the same for $|y|$.

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