Evaluating Limit of $\frac{xyz}{x^2 + y^2 + z^2}$ using $\epsilon$ - $\delta$ method I've been working on this for a little while.
I understand the way to do this is using spherical coordinates and the limit = 0, but I'm also wondering if it's possible to solve it using the epsilon-delta method.
Here is some of what I have:
Since I know that
$$\lim_{(x,y,z) \to (0,0,0)} \frac{xyz}{x^2 + y^2 + z^2} = 0$$
I start with
$$\left|\frac{xyz}{x^2 + y^2 + z^2}\right|<\epsilon$$
and
$$\sqrt{x^2 + y^2 + z^2} < \delta$$
Which seems to indicate
$$x < \delta \quad ; \quad y < \delta \quad ; \quad z < \delta$$
And thus
$$ \frac{\delta ^3}{x^2 + y^2 + z^2} < \epsilon$$
Since
$$ x^2 + y^2 + z^2 < \delta^2$$
This implies
$$ \frac{\delta^3}{\delta^2} < \frac{\delta ^3}{x^2 + y^2 + z^2} < \epsilon$$
And finally
$$ \delta < \epsilon$$
Which is fine.. but if this is the "game" that some people use to explain the method, I've only claimed that the $\delta$ I provide will be smaller than the $\epsilon$ provided. Am I interpreting this correctly, or does this mean: "any number larger than $\sqrt{x^2 + y^2 + z^2}$ will be smaller than $\epsilon$"? Which I suppose means any finite number > 0 will work.
From the examples that I've seen in my book and online, they usually set $\delta = g(\epsilon)$ by solving for an inequality like $\sqrt{x^2+y^2+z^2} < g(\epsilon)$ using the inequality that starts like $|f(x) - L| < \epsilon$. I can't feel but think I made some mistake.
I appreciate any help!

The idea is: given $\varepsilon>0$, to find a $\delta>0$ such that $\sqrt{x^2+y^2+z^2}<\delta\implies\left|\frac{xyz}{x^2+y^2+z^2}\right|<\varepsilon$. So, fix some $\varepsilon>0$. Now, let $r=\sqrt{x^2+y^2+z^2}$. Then $|x|,|y|,|z|\leqslant r$ and therefore$$\left|\frac{xyz}{x^2+y^2+z^2}\right|\leqslant\frac{r^3}{r^2}=r.$$So, take $\delta=\varepsilon$. This will work, because if $\sqrt{x^2+y^2+z^2}=r<\delta=\varepsilon$, then $\left|\frac{xyz}{x^2+y^2+z^2}\right|\leqslant r<\varepsilon$.
