Using $\epsilon$-$\delta$ approach to prove that $\lim_{(x,y,z)\rightarrow (0,0,0)}\frac {y^3-1000xy^2+z^5}{x^2+y^2+z^4}=0$? As the title indicates, Using $\epsilon$-$\delta$ approach to prove that $$\lim_{(x,y,z)\rightarrow (0,0,0)}\frac {y^3-1000xy^2+z^5}{x^2+y^2+z^4}=0$$?
 A: Here is a method without spherical coordinates, only crude inequalities. First, we have:
$$\left| \frac{y^3-1000xy^2+z^5}{x^2+y^2+z^4} \right| \leq \frac{|y|^3}{x^2+y^2+z^4}+1000\frac{|x|y^2}{x^2+y^2+z^4}+\frac{|z|^5}{x^2+y^2+z^4}.$$
Since each of $x^2$, $y^2$ and $z^4$ is non-negative, we have $x^2+y^2+z^4 \geq y^2$ and $x^2+y^2+z^4 \geq z^4$, so that:
$$\left| \frac{y^3-1000xy^2+z^5}{x^2+y^2+z^4} \right| \leq \frac{|y|^3}{y^2}+1000\frac{|x|y^2}{y^2}+\frac{|z|^5}{z^4} = |y|+1000|x|+|z|.$$
Hence, the limit at $(0,0,0)$ of this function is $0$.
A: Let
$$
\begin{eqnarray}
x &=& r\cos\phi\cos\theta &=& rau &\qquad& a &=& \cos\phi  &\qquad& u &=& \cos\theta \\
y &=& r\cos\phi\sin\theta &=& rav &\qquad& b &=& \sin\phi  &\qquad& v &=& \sin\theta \\
z &=& r\sin\phi           &=& rb
\end{eqnarray}
$$
Then we have
$$
\lim_{r \to 0} \frac{r^3a^3v^2(v-1000u)+r^5b^5}{r^2a^2+r^4b^4} =
\lim_{r \to 0} \frac{r  a^3v^2(v-1000u)+r^3b^5}{   a^2+r^2b^4} = 0
$$
which vanishes everywhere (and is thus well-defined).
To see this, consider first that when $a\neq0$,
the denominator is dominated by the $a^2$ term
so that the limit becomes
$$
\lim_{r \to 0} \; rav^2(v-1000u) = 0
$$
while when $a=0$ (i.e. at the north and south poles where $b=\pm1$),
the limit becomes
$$
\lim_{r \to 0} \frac{r^3b^5}{r^2b^4} = 
\lim_{r \to 0} \; rb = 0
$$
which is still independent of the direction of approach to the origin.
A: If you choose spherical coordinates and fiddle with trigonometric inequalities you should obtain 0... Am I missing something?
