Limit as $(x,y,z)\to (0,0,0)$ of $f(x,y,z) = \dfrac{xy+yz+xz}{\sqrt{x^2+y^2+z^2}}$

Question

To find this limit, I converted to spherical coordinates and rewrote:

$$\lim_{r\to 0} \dfrac{r^2(\sin^2\theta \cos\phi \sin \phi + \sin\theta \cos \theta \sin \phi + \sin\theta \cos \theta \cos \phi)}{r} = 0$$

Is this method alright? Our teacher did using epsilon delta proof, so how can we use something similar to spherical coordinates if say we had four variable limit of kind:

$$\lim_{(w,x,y,z) \to (0,0,0,0)} \frac{xy+yz+xz+wx}{ \sqrt{x^2+y^2+z^2+w^2}}$$

Answer 1

Using the full spherical coordinates is overkill here. Let $r=\sqrt{x^2+y^2+z^2}$. Then $|x|\le r$, $|y|\le r$, $|z|\le r$. So $$|xy+xz+yz|\le|xy|+|xz|+|yz|\le 3r^2$$ and so $$\left|\frac{xy+xz+yz}{\sqrt{x^2+y^2+z^2}}\right|\le 3r.$$ As $\lim_{(x,y,z)\to(0,0,0)}r= 0$ then $$\lim_{(x,y,z)\to(0,0,0)}\left|\frac{xy+xz+yz}{\sqrt{x^2+y^2+z^2}}\right|=0$$ also.

This method works for your four-variable problem too, avoiding the minutiae of four-dimensional spherical coordinates.

Answer 2

It can be done. Let $$r = \sqrt{x^2+y^2+z^2+w^2}$$ $$x = r\cos(\phi_1)$$ $$y = r\sin(\phi_1)\cos(\phi_2)$$ $$z = r\sin(\phi_1)\sin(\phi_2)\cos(\phi_3)$$ $$w = r\sin(\phi_1)\sin(\phi_2)\sin(\phi_3).$$ Note that we transformed the coordinates $(x,y,z,w)$ to spherical coordinates $(r,\phi_1,\phi_2,\phi_3)$ with $\phi_1,\phi_2$ ranging over $[0,\pi]$ and $\phi_3$ over $[0,2\pi].$ Then notice that you will get the same simplification as in the three dimension case.

Answer 3

Notice that $x,y$ and $z$ are less and equal than $\sqrt{x^2+y^2+z^2}$ so that

$$ |xy+yz+xz| \leq 3\left(\sqrt{x^2+y^2+z^2}\right)^2.$$

Hence, by a straightforward application of sandwich theorem, there holds that the aforementioned limit equals zero, since

$$ 0\leq \lim_{(x,y,z)\rightarrow (0,0,0)}\left|\dfrac{xy+yz+xz}{\sqrt{x^2+y^2+z^2}}-0 \right|\leq \lim_{(x,y,z)\rightarrow (0,0,0)}3\sqrt{x^2+y^2+z^2}=0.$$