Finding multivariable limit $\lim_{(x,y,z)\to(0,0,0)} \frac{xyz}{(x^2+y^2+z^2)^{a/2}}$ Let $a>0$. Find $$\lim_{(x,y,z)\to(0,0,0)} \frac{xyz}{(x^2+y^2+z^2)^{a/2}}$$
After playing around with this a little bit, it looks like the limit is $0$ for $a<3$ and else it is $\infty$. But how do I actually compute this limit? Are there any nice tricks to evaluate limits of three variables?
Thank you.
 A: $|\frac{xyz}{(x^2+y^2+z^2)^{a/2}} |\leq \frac{r^3}{r^a} = r^{3-a}$ where $r$ is your radius in spherical coordinates, ie $M(x,y,z) = M(r,\theta,\phi)$ with $OM=r$.
If $a<3$ then the above limit converges to 0 as show by the inequation above.
If $a\geq3$ then it will depend on the path you use, as explained in the other answers.
This is always how this kind of exercise works: use $r$ to bound the function and you're back to a simple limit. Then find 2 sequences that do not converge to the same value.
A: Sure. I checked my answer on wolfram to verify it too.
So you want to turn the expression into 
$\frac {{\rho^3} cos (\theta) sin(\theta) {sin (\phi)^2} cos(\phi)}{\rho^(2a/2)} $
Do you see where this comes from?
This simplifies to
${\rho^{3-a}} cos (\theta) sin (\theta){sin (\phi)^2} cos (\phi) $
The limit becomes $(\rho,\phi,\theta) \mapsto (0, \pi,2\pi)$. Obviouslt the $\rho=0$ makes this limit zero no matter what line/curve we approach the the center at thus the limit must be zero.
HOWEVER this is for $ a <3$ as you said. $a> 3$ we get divide by zero hence Undefined limit.
At $ a=3$  it is also undefined. $ a=3$ mean $\rho $ is always 1. In the above case I said the two angle were $\pi $ ans $2\pi $, but they don't have to be. This angles dont make the function approach the origin, so they can be whatever they want. Thus with $\rho $ being one you get all the trig junk above which can never approach the origin in spherical coordinated, and thus the limit doesn't exist for $ a=3$ 
You can also do this in cartesian as suggested below. But in cartesian you need select more paths and possibly use the squeeze theorem.
A: Let $x=y=x$ therefore we have
$$\lim\limits_{(x,y,z)\to(0,0,0)}\frac{xyz}{(x^2+y^2+z^2)^\frac{a}{2}}=\lim\limits_{x\to0}\frac{x^3}{(3x^2)^\frac{a}{2}}=\frac{1}{\sqrt[a]{3}}\lim\limits_{x\to0}\frac{x^3}{x^{a}}\ \ \ (1)$$ and if we consider $x=y$ and $z=0$ then 
$$\lim\limits_{(x,y,z)\to(0,0,0)}\frac{xyz}{(x^2+y^2+z^2)^\frac{a}{2}}=0  \ \ \ \ (2)$$ so that from (1) and (2)  we see that if $a\geq3$ limit don't exits
