Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
1  
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 –  sidht 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. –  sidht Feb 5 '13 at 22:31

2 Answers 2

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 $0$.

share|improve this answer
    
Does it really matter if $r \to 0^+$ or $-$? –  sidht 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 $r^2 = x^2 + y^2$ –  sidht 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 $r$. –  Arkamis 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 $|y|$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.