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How do I prove that $$\lim_{z\to0}\frac{\Re(z)\Im(z)}{|z|}=0? $$ I've tried it using $\varepsilon-\delta$ language, but can't really get anywhere.

Any hints would be greatly appreciated.

Thanks.

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    The same limit came up in another question: http://math.stackexchange.com/questions/26672/finding-limit-of-function-with-more-than-one-variable2011-04-16

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Clearly neither Re(z) nor Im(z) can have magnitude larger than |z|. So:

$\displaystyle{ \lim_{z\to0}\left|\frac{\mathrm{Re}(z)\mathrm{Im}(z)}{|z|}\right| = \lim_{z\to0}\frac{|\mathrm{Re}(z)|\times|\mathrm{Im}(z)|}{|z|} \le \lim_{z\to0}\frac{|z|\times|z|}{|z|} = \lim_{z\to0}|z| = 0 }$

Since the magnitude goes to zero, the quantity goes to zero.

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If we rewrite it

$$\lim_{(x, y) \to (0, 0)} \frac{xy}{\sqrt{x^2+y^2}}$$

does it look more familiar? You can use polar coordinates or Young's inequality.

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    So if we write x = rcos(t), y = rsin(t), we have r^2sin(t)cos(t) / r = rsin(t)cos(t). What is tending to 0 now? Is it r, for any fixed theta?2011-04-16
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    Yes. But, please, use the correct syntax for formulas. Have a look at the FAQ located in the upper part of this page. EDIT: When $r$ tends to $0$ we have $r \sin(\theta)\cos(\theta) \to 0$ uniformly with respect to $\theta \in [0, 2\pi]$. This means that the original expression tends to $0$ when $(x, y) \to (0,0)$.2011-04-16
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    Okay, thank you for your help (and for directing me to the FAQs).2011-04-16
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    good proof of course, but just for reference, do you think an inequality-based proof exists that does not rely on Young's? I think a lot of undergrads see Young's ineq. relatively late, at least after they've done complex analysis.2011-04-16
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    @Gerben: To say the truth it's just my laziness that made me mention Young's inequality. The inequality you need here is $|xy|\le \frac{1}{2}(|x|^2+|y|^2)$ which you can prove in a nanosecond by observing that $(\lvert x \rvert-\lvert y \rvert)^2\ge 0$. Sure, it is a special case of Young, but it's much easier.2011-04-16