# Show that $\lim_{(x,y)\to (0,0)} {xy\over \sqrt{x^2+y^2}}$ exists

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?

• Try polar coordinates. – saulspatz Aug 2 '19 at 16:54

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.

• I do not quite understand why the LHS less than x can conclude the limit exists. – Brian Wu Aug 2 '19 at 17:12
• 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. – Peter Foreman Aug 2 '19 at 17:15
• 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. – Lutz Lehmann Aug 2 '19 at 19:28
• @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? – Nzewi Ernest Kenechukwu Aug 3 '19 at 13:51

Polar coordinate gives you $$\lim_{(x,y)\to (0,0)} {xy\over \sqrt{x^2+y^2}}=\lim_{r\to 0} {r^2 \sin \theta \cos \theta\over r} =$$

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

• Is it enough to prove that the limit exists? – Brian Wu Aug 3 '19 at 4:52
• Yes, it shows that limit exists and it is $0$u – Mohammad Riazi-Kermani Aug 3 '19 at 6:14

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 $$0$$.