Limit of given 3 variable function

Question

I am trying to find the limit for this function at (0,0,0) and I am getting 0 as the limit for 4 different paths I tried. So I think 0 is the limit but how do I prove it?

$f(x,y,z) = {x^2 y^2 z^2\over x^2 + y^2 + z^2}$.

Answer 1

For $x=0,$ we have $ f(x,y,z)=0.$

Otherwise. $$ 0 \le f(x,y,z) = {x^2 y^2 z^2\over x^2 + y^2 + z^2}\le {x^2 y^2 z^2\over x^2 } =y^2 z^2 \to 0$$

Answer 2

You can also use the polar coordinates $$ x=r\sin(\theta)\cos(\psi),~y=r\sin(\theta)\sin(\psi), z=r\cos(\theta). $$ Then we get $|(x,y,z)|=r$ and $$ |f(x,y,z)|=\left|\frac{r^2\sin^2(\theta)\cos^2(\psi)r^2\sin^2(\theta)\sin^2(\psi)r^2\cos^2(\theta)}{r^2}\right|\leq r^4. $$

Now, you can use the $\epsilon-\delta$-definition to proof continuity of $f$ at $(0,0,0)$.

You can see it here:

Let be $\epsilon>0$ and define $\delta=\epsilon^{1/4}$. For all $(x,y,z)\in\mathbb R^3$ such that $|(x,y,z)-(0,0,0)|<\delta$ holds, using the polar coordinates, $r=|(x,y,z)|<\delta$ and $$|f(x,y,z)-f(0,0,0)|=|f(x,y,z)|\leq r^4<\delta^4=\epsilon.$$

Answer 3

AM-GM:

$x^2+y^2+z^2 \ge 3 \sqrt[3]{x^2y^2z^2}$; or

$(x^2+y^2+z^2)^3 \ge 3^3 (x^2y^2z^2).$

$|f(x)-0| =\dfrac{x^2y^2z^2}{x^2+y^2+z^2} \le $

$\dfrac{(x^2+y^2+z^2)^3}{3^3(x^2+y^2+z^2)} \lt $

$(x^2+y^2+z^2)^2.$

Let $\epsilon >0$ be given.

Choose $\delta = \sqrt[4]{\epsilon}$, then

$ |x^2+y^2+z^2|^{1/2} \lt \delta$ implies

$|f(x)| \lt (x^2+y^2+z^2)^2 \lt \delta^4 = \epsilon$.