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Find the limit $$\lim_{(x,y)\to (0,0)} {xy\over \sqrt{x^2+y^2}}$$

By approaching the origin along both $x,y$-axis, I got the same result $0$.

So how can I prove the limit exists by epsilon-delta definition?

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    $\begingroup$ Try polar coordinates. $\endgroup$ – saulspatz Aug 2 '19 at 16:54
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Note that we have $$\left|\frac{xy}{\sqrt{x^2+y^2}}\right|\lt\left|\frac{xy}{\sqrt{y^2}}\right|=|x|$$ hence the limit exists and equals zero.

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  • $\begingroup$ I do not quite understand why the LHS less than x can conclude the limit exists. $\endgroup$ – Brian Wu Aug 2 '19 at 17:12
  • $\begingroup$ Because as long as $|x|\lt\epsilon$ we can ensure that the given function is also $\lt\epsilon$ so by the $\epsilon-\delta$ definition, the limit exists and is zero. $\endgroup$ – Peter Foreman Aug 2 '19 at 17:15
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    $\begingroup$ Or use the more symmetric $x^2+y^2\ge\max(|x|,|y|)^2$ to get $$\left|\frac{xy}{\sqrt{x^2+y^2}}\right|\le \min(|x|,|y|)$$ which is less singular outside the origin. $\endgroup$ – Lutz Lehmann Aug 2 '19 at 19:28
  • $\begingroup$ @Lutzl, your approach seems very much reasonable for me in understanding the problem at hand. If I may ask, what text did you get the concept from? $\endgroup$ – Nzewi Ernest Kenechukwu Aug 3 '19 at 13:51
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Polar coordinate gives you $$\lim_{(x,y)\to (0,0)} {xy\over \sqrt{x^2+y^2}}=\lim_{r\to 0} {r^2 \sin \theta \cos \theta\over r} =$$

$$\lim _{r\to 0} r\sin \theta \cos \theta =0$$

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  • $\begingroup$ Is it enough to prove that the limit exists? $\endgroup$ – Brian Wu Aug 3 '19 at 4:52
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    $\begingroup$ Yes, it shows that limit exists and it is $0$u $\endgroup$ – Mohammad Riazi-Kermani Aug 3 '19 at 6:14
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Because of $0\le(|x|-|y|)^2$ you get $2|xy|\le x^2+y^2$. Thus $$ \frac{|xy|}{\sqrt{x^2+y^2}}\le\frac{\sqrt{x^2+y^2}}2 $$ which proves continuity in $(x,y)=(0,0)$ with value $0$.

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