# How do I find the limit of $\frac{xy\sqrt{|xy|}}{x^2 + xy + y^2}$ as x and y approach zero?

I am trying to find: $$\lim_{(x,y)\to (0,0)}\frac{xy\sqrt{|xy|}}{x^2 + xy + y^2}$$

I suspect that the limit does exist as the combined power of $x$ and $y$ is higher in the numerator than in the denominator, and I have noticed a pattern where this produces a limit, but the reverse case does not.

I have tried using polar coordinates $x = r\cos{\theta}, y = r\sin{\theta}$ and simplifying to get:

$$f(r\cos{\theta}, r\sin{\theta}) = \frac{r(\cos{\theta}\sin{\theta})\sqrt{|\cos{\theta}\sin{\theta}|}}{\cos{\theta}\sin{\theta} + 1}$$

I can't seem to get anywhere from here. I was trying to apply the squeezed theorem, but this expression seems like it needs to be simplified more before I can do that. Any hints on how to do that? Or am I barking up the wrong tree and need to try another approach?

Many thanks for any insights :)

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## 2 Answers

Since $\cos\theta\sin\theta=\frac 12\sin 2\theta$, $1+\cos\theta\sin\theta\geq \frac 12$ so $|f(r\cos\theta,r\sin\theta)|\leq 2r$ and you are done.

Alternatively, we can write $x^2+y^2\geq 2|xy|$ so $x^2+y^2+xy\geq 2|xy|+xy\geq |xy|$ and for $xy\neq 0$ we have $|f(x,y)|\leq \sqrt{|xy|}$.

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Aaah double angle formulae, why do I neglect you? Thanks! – Katherine Rix Mar 5 '12 at 8:33

Looking at your $$f(r\cos\theta, r\sin\theta) = \frac{r \cos\theta\sin\theta\sqrt{|\cos\theta\sin\theta|}}{\cos\theta\sin\theta + 1}$$ one immediately sees that $\bigl|f(r\cos\theta, r\sin\theta)\bigr|\leq r$, whence $$\bigl|f(x,y)\bigr|\leq\sqrt{x^2+y^2}\ ,$$ and the requested limit is $0$.

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