$\lim_{(x,y,z) \to (0,0,0)} \frac{xyz}{x^2+y^2+z^2}=0$

Question

How to show that $$\lim_{(x,y,z) \to (0,0,0)} \frac{xyz}{x^2+y^2+z^2}=0,$$ where $x,y,z>0$.

My attempt:

$$||(x,y,z)|| < \delta \implies |x|, |y|, |z| < \delta$$ $$\left | \frac{xyz}{x^2+y^2+z^2} \right | < \left | \frac{xyz}{x^2}\right | < \frac{\delta^3}{x^2}.$$

Now, I do not know how to proceed, and I think my attempt might be wrong.

Mark Viola · Accepted Answer · 2020-09-17T02:53:17.260

3

Note that $|x|\le \sqrt{x^2+y^2+z^2}$, $|y|\le \sqrt{x^2+y^2+z^2}$, and $|z|\le \sqrt{x^2+y^2+z^2}$.

Hence, given $\varepsilon>0$,

$$ \left| \frac{xyz}{x^2+y^2+z^2}\right|\le \sqrt{x^2+y^2+z^2}<\varepsilon$$

whenever $\sqrt{x^2+y^2+x^2}<\delta=\varepsilon$.

And we are done!

edited Sep 17 '20 at 02:53

answered Sep 17 '20 at 02:42

Mark Viola

179,405

score 3 · Answer 2 · answered Sep 17 '20 at 02:44

3

Using A-G inequality, $x^2+y^2+z^2\geq3(xyz)^{\frac{2}{3}}$

answered Sep 17 '20 at 02:44

user823011

1,335

1

A more general way is let $x=r\sin\varphi\cos\theta, y=x=r\sin\varphi\sin\theta, z=r\cos\theta$, then let $r\rightarrow 0$. – user823011 Sep 17 '20 at 02:47
2

@user8w3011 Use of polar coordinates is completely unnecessary. – Mark Viola Sep 17 '20 at 02:49
I think it's a commonly-used and convenient way to solve limits of rational functions like this, maybe this is just my opnion. – user823011 Sep 17 '20 at 02:58

$\lim_{(x,y,z) \to (0,0,0)} \frac{xyz}{x^2+y^2+z^2}=0$

2 Answers2