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

Question

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?

Answer 1

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.

Answer 2

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

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

Answer 3

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 $0$.