Existence of $\lim_{(x,y,z)\to(0,0,0)}\frac{\ln(1+x^2y^4z^6)}{(x^2+y^2+z^2)^{\alpha+\frac{1}{2}}}$ for some $\alpha$. Hy guys!
for which values of $\alpha\in\mathbb R$ does this limit exists? 
$$\lim_{(x,y,z)\to(0,0,0)}\frac{\ln(1+x^2y^4z^6)}{(x^2+y^2+z^2)^{\alpha+\frac{1}{2}}}$$
I know that for all $\alpha\leq-\frac{1}{2}$ the limit exists, but how can I argue to prove (or disprove) that this limit exists for $\alpha>-\frac{1}{2}$?
 A: We use spherical coordinate system for solving this limit. In the spherical coordinate system we suppose that
$$
\begin{array}{l}
x=r\,\sin(\theta)\,\cos(\phi)\\
y=r\,\sin(\theta)\,\sin(\phi)\\\tag{1}
z=r\, \cos(\theta)
\end{array}
$$
and the relation between $x$,$y$ and z is as follows 
$$
x^2+y^2+z^2=r^2\tag{2}
$$
If we substitute the values of  $x$,$y$ and z by $(1)$ when $r$ tends to zero and the value of limit exist and be independent of $\theta$and $\phi$, we say the limit exists and in the other cases we conclude that the limit dose not exist. So by using $(1)$ and $(2)$, we have 
$$
\lim_{(x,y,z)\to(0,0,0)}\frac{\ln(1+x^2y^4z^6)}{(x^2+y^2+z^2)^{\alpha+\frac{1}{2}}}=
\lim_{r\to 0}\frac{\ln(1+{(r\,\sin(\theta)\,\cos(\phi))}^2\,{(r\,\sin(\theta)\,\sin(\phi))}^4\,{(r\, \cos(\theta))}^6)}{(r^2)^{\alpha+\frac{1}{2}}}
$$
$$
=\lim_{r\to 0}\frac{ln(1+r^{12}\,\sin(\theta)^{12}\,\cos(\phi)^{12})}{r^{2\alpha+1}}=\lim_{r\to 0} \frac{ln(1)}{r^{2\alpha+1}}\tag{3}
$$
In $(3)$ we use from this fact that limit of zero at the bounded functions is zero. The limit $(3)$ exist if $2\alpha+1\leq0$ or $\alpha\leq -\frac{1}{2}$ and for $\alpha> -\frac{1}{2}$ dose not exist. 
I hope you find it useful. 
