Prove that the limit as $(x,y,z)$ approaches $0$ of $(x^3+y^3+z^3)/(xyz)$ does not exist I am asked to show that 
$$
\lim_{(x,y,z)\to(0,0,0)}\frac{x^3+y^3+z^3}{xyz}
$$
does not exist. I know that I need to use $x=at$, $y=bt$, and $z=bt$ but I don't understand what I should do after inserting them into the function.
 A: For these types of limits, think of a point in the space. Now, you will grab that point and make it go to the origin, by a path of your liking. That will make the quotient in the limit take a specific value. If you find two paths that give a different value of the limit, you proved the limit does not exist.
As a practical example, take
$$\lim_{x,y\to 0} \frac{x}{y} $$
Does that limit exist?
I am thinking of picking a point in the line $y = x $, and travelling on top of it all the way to the origin. For that particular path, the limit equals $1$. Can you see why?
Now, instead of going to the origin by the line $y = x $, I'll take $y = -x $. With that, I will have that the limit equals $-1$. Can you see why?
When we suspect a limit divirges, it is often worthwhile trying to take $y = mx $, and then simplifying the limit. If the final expression depends on $m $, the limit is not unique for every path, as picking two different lines with different slopes as paths gives different values.
We are applying that technique, but with $3$ variables. Try writing $at = x = y = z $ or $y = z = mx $ and simplify the expression inside the limit. If the limit of the final expression depends on your parameters, the limit does not exist.
A: The fact that $xyz$ equals $0$ on the axes while $x^3+y^3+z^3$ does not suggests that a path close to one of the axes will give bad behavior. For example, consider the path $(x,x^3,x^3)$ as $x\to 0.$ Along this path our function equals
$$\frac{x^3 + 2x^9}{x^7}=  \frac{1}{x^4} + 2x^2.$$
This $\to \infty$ as $x\to 0.$ Along the path $(x,-x^3,x^3)$ we get the limiting value $-\infty.$
A: In
$\frac{x^3+y^3+z^3}{xyz}
$
let
$y = ax$
and
$z = bx$.
These values are on
a line through the origin.
It becomes
$\frac{x^3+(ax)^3+(bx)^3}{x(ax)(bx)}
=\frac{x^3+a^3x^3+b^3x^3}{abx^3}
=\frac{1+a^3+b^3}{ab}
$.
Therefore,
along this line,
the ratio depends
only on $a$ and $b$.
Therefore
the limit does not exist
because,
by choosing different
$a, b$ values,
the ration can take on
different values.
For example,
if $a=b=1$,
the ratio is $3$;
if $a=b=2$,
the ratio is $17/4$;
if $b=a$,
the ratio is $(1+2a^3)/a^2$;
and so on.
