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Hey I need to show that $$\lim_{(x,y)\to(0,0)} {xy \over \sqrt{x^2 + y^2}} = 0.$$ I'm not sure how to start manipulating this as I haven't gotten anything useful yet. Some help to get me going would be nice. Thanks

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up vote 4 down vote accepted

Notice that $$|x| = \sqrt{x^2} \le \sqrt{x^2 + y^2}$$ Using this fact, we have $$\left\vert\frac{xy}{\sqrt{x^2 + y^2}}\right\vert = \frac{|x|}{\sqrt{x^2 + y^2}}|y|\le |y|$$ which approaches $0$ as $(x,y)\rightarrow 0$.

It should not be too hard to convert this into a formal epsilon-delta proof.

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Let $$f(x,y) = \dfrac{xy}{\sqrt{x^2 + y^2}}$$ Set $x = r \cos(\theta)$, $y = r \sin(\theta)$, let $\sqrt{x^2 + y^2} = r$. $$F(r,\theta) = f(r \cos(\theta), r \sin(\theta)) = \dfrac{r^2 \cos(\theta) \sin(\theta)}{r} = r \cos(\theta) \sin(\theta)$$ $$(x,y) \to (0,0) \implies r \to 0$$ Now can you conclude what you want?

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Yes, but this is not using epsilon-delta – William Sep 26 '12 at 2:01
@William: But this is a very formal approach (+1). :) – Babak S. Sep 26 '12 at 6:06

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