Correct reasoning for continuity of $f$ in $(0,0,0)$

Question

Let $f: \mathbb R^3-\{0\}\to \mathbb R, f(x,y,z)=\frac{x^3+2y^3-z^2}{x^2+y^2+z^2}$.

Can the continuity of $f$ be extended to $(0,0,0)$? I would like to yes, but I am not sure whether my reasoning is sound.

My ideas:

$\frac{x^3+2y^3-z^2}{x^2+y^2+z^2}=\frac{x^3}{x^2+y^2+z^2}+\frac{2y^3}{x^2+y^2+z^2}-\frac{z^2}{x^2+y^2+z^2}$

Now for $\vert\frac{x^3}{x^2+y^2+z^2}\vert\leq\vert\frac{x^3}{x^2}\vert=\vert x \vert\to0$, as $x\to0$

Similarly $\vert\frac{2y^3}{x^2+y^2+z^2}\vert\leq2|y|\to y$, as $y \to 0$

And then

$\vert\frac{z^2}{x^2+y^2+z^2}\vert\leq\vert\frac{z^2}{z^2}|=1$

Does this show $f$ is continuous in $0$. Similarly does it then show $f(0)=1$? Or does it rather show $\exists c \in \mathbb R^{3}$ such that $f(0)=c?$

Answer 1

What happens when we take $f(0,0,\frac1n)$ when $n \to \infty$? And what with $f(\frac1n, 0,0)$ as $n \to \infty$. If $f$ could be extended continuously to $\mathbb{R}^3$, both would have to converge to $f(0,0,0)$.

Answer 2

Hint

Compute $$\lim_{t\to 0} f(t,0,0)\quad \text{and}\quad \lim_{t\to 0} f(0,0,t).$$