squeeze theorem and 3 variable function's limit

Question

$f(x,y,z) = \frac{xyz}{x^2+y^2+z^2}$. Its limit at $(0,0,0)$ is $0$ and we had to prove it. My TA in tutorial did it this way: $\left|\frac{xyz}{x^2+y^2+z^2}\right|\leq \frac{x^2+y^2+|xyz|}{x^2+y^2+z^2}$

and since $\frac{x^2 + y^2}{x^2+y^2+z^2} \leq 1$ so, $\frac{x^2+y^2+|xyz|}{x^2+y^2+z^2}\leq|xyz|$ which tends to $0$ at $(0,0,0)$ so limit is $0$.

But isn't this wrong? Shouldn't this work out only if $x^2+y^2$ and $|xyz|$ were multiplied to each other? Can someone tell me what should be the correct method.

Answer 1

Yes, you're right that the TA was in error. He confused multiplication and addition.

You most definitely do not need to think about spherical coordinates. (Generally, people on MSE resort to polar coordinates and spherical coordinates when it's not necessary.) Note that the key idea is that $$|x|, |y|, |z|\le \sqrt{x^2+y^2+z^2}$$ (and this works in however many dimensions you wish). So $$0\le \frac{|xyz|}{x^2+y^2+z^2} \le \frac{(x^2+y^2+z^2)^{3/2}}{x^2+y^2+z^2} = \sqrt{x^2+y^2+z^2},$$ and now we can easily apply the squeeze theorem to deduce the limit is $0$.

Answer 2

Yes you are right the solution given is wrong, by inequality we can proceed as shown bt Ted Shifrin or as an alternative by AM-GM

$$\frac{x^2+y^2+z^2}{3}\ge\sqrt[3]{(xyz)^2}\implies |xyz|\le\frac{\sqrt{(x^2+y^2+z^2)^3}}{3\sqrt{3}}<(x^2+y^2+z^2)\sqrt{x^2+y^2+z^2}$$

then

$$\left|\frac{xyz}{x^2+y^2+z^2}\right|= \frac{|xyz|}{x^2+y^2+z^2}<\sqrt{x^2+y^2+z^2}\to 0$$

Inequality are of course a good and very elegant method to proceed but not always those way are simple to handle (indeed your TA made a mistake).

Thus as a good alternative we can also solve by spherical coordinates

• $x=r\sin \phi \cos \theta$
• $y=r\sin \phi \sin \theta$
• $z=r\cos \phi$

for $r=\sqrt{x^2+y^2+z^2}\to 0$ we have

$$\frac{xyz}{x^2+y^2+z^2}=r\cdot f(\theta,\phi) \to 0$$

since $f(\theta,\phi)=\sin^2 \phi\cos \phi \cos \theta\sin \theta$ is bounded.