Vtyjkyuk

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

$$lim_rto 0 dfracr^2(sin^2theta cosphi sin phi + sintheta cos theta sin phi + sintheta cos theta cos phi)r = 0$$

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:

$$lim_(w,x,y,z) to (0,0,0,0) fracxy+yz+xz+wx
sqrtx^2+y^2+z^2+w^2$$

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

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

It can be done. Let
$$r = sqrtx^2+y^2+z^2+w^2$$
$$x = rcos(phi_1)$$
$$y = rsin(phi_1)cos(phi_2)$$
$$z = rsin(phi_1)sin(phi_2)cos(phi_3)$$
$$w = rsin(phi_1)sin(phi_2)sin(phi_3).$$
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 $[0,2pi].$
Then notice that you will get the same simplification as in the three dimension case.

Notice that $x,y$ and $z$ are less and equal than $sqrtx^2+y^2+z^2$ so that

$$ |xy+yz+xz| leq 3left(sqrtx^2+y^2+z^2right)^2.$$

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

$$ 0leq lim_(x,y,z)rightarrow (0,0,0)left|dfracxy+yz+xzsqrtx^2+y^2+z^2-0 right|leq lim_(x,y,z)rightarrow (0,0,0)3sqrtx^2+y^2+z^2=0.$$