Limits (Three Variable function). We're given : 
$f(x,y,z) = \dfrac{xyz}{x^{2}+y^{2}+z^{2}}$ ,
Also , it's given that $\lim_{(x,y,z) \to (0,0,0)} f(x,y,z)$ exists.
We need to prove that $\lim_{(x,y,z) \to (0,0,0)} f(x,y,z) = 0$.
What I figured : 
First I approach $(0,0,0)$ along $x$ -axis , and thus the limit becomes :
$\lim_{(x,y,z) \to (0,0,0)\dfrac{x.0.0}{x^{2}+0^{2}+0^{2}}}$ which is $0$.
Similarly approaching along $y$ and $z$ axis gives the limit to be $0$.
Now , since the limit at $(0,0,0)$ exists , it should be unique throughout , thus , 
$\lim_{(x,y,z) \to (0,0,0)} f(x,y,z) = 0$
Could anyone tell , am I right in making the above statement ?
 A: Yes. But what I don't understand is why you were told that the limit exists. It is a totally redundant assumption.
Let $r = \sqrt{x^2+y^2+z^2}$.
Then $|x| \le r$ and $|y| \le r$ and $|z| \le r$.
Thus $|xyz| \le r^3$ and hence $\left| \frac{xyz}{x^2+y^2+z^2} \right| \le r \to 0$ as $r \to 0$.
Therefore $\lim_{(x,y,z) \to (0,0,0)} \frac{xyz}{x^2+y^2+z^2} = \lim_{r \to 0} \frac{xyz}{x^2+y^2+z^2} = 0$.
A: Yes, if the limit exists it's unique and you are correct. However you only need to check one particular direction, so just checking the $x$ axis would be enough 
A: Little observation: in general (if you do not know whether the limit exists), if the limit exists along one path it does NOT mean that it exists, so your proof for $(x,0,0)$ is not sufficient. 
For the general case, therefore, I would either use the demonstration of answer $1$ , or majorize at the numerator $|xy|$ by $\frac{x^2+y^2}{2}$, and minorize at the denominator $x^2+y^2+z^2$ by $x^2+y^2$, in order to get that $0\leq |f(x,y,z)|\leq \frac{|z|}{2}\leq\frac{\sqrt{x^2+y^2+z^2}}{2}$, which goes to zero for $(x,y,z)$ going to $(0,0,0)$.
For the sandwich theorem, the limit of your function at the origin is zero.
