Limit of $\frac{2n}{ \left( \log{(n^{10} m)} \right)^m }$ I'm trying to prove
$$\displaystyle \lim_{n \to \infty}\frac{2n}{ \left( \log{(n^{10} m)} \right)^m } = \begin{cases}
    0       & \quad \text{if } m=\log{n} / \left( \left(\log{\log{n}}\right)^1 \right)\\
    \infty  & \quad \text{if } m=\log{n} / \left( \left(\log{\log{n}}\right)^2 \right)
  \end{cases}$$
This is a modification of this question.  I think that this version will hopefully be doable.
I got the limits through Mathematica 9.0, so they should be correct.
 A: With
$$m = \frac{\log n}{(\log \log n)^{\alpha}}$$
we find the behaviour of
$$\frac{An}{\bigl(\log (n^Bm)\bigr)^m}$$
as $n \to \infty$, for arbitrary positive $A,B,\alpha$ easily by taking the logarithm. This yields
\begin{align}
\log A + \log n &{}- m\bigl(\log \log (n^Bm)\bigr) \\
&= \log A + \log n - \frac{\log n}{(\log \log n)^{\alpha}}\log (B\log n + \log m) \\
&= \log A + \log n - \frac{\log n}{(\log \log n)^{\alpha}}\biggl(\log B + \log \log n + \log\biggl(1 + \frac{\log m}{B\log n}\biggr)\biggr).
\end{align}
Now for $\alpha > 1$ it is easy to see that the term we subtract belongs to $o(\log n)$, so the limit is $+\infty$ for $\alpha > 1$. For $\alpha = 1$, the $\log n$ from the numerator is cancelled by the second term in the parenthesis, and since $\frac{\log n}{\log \log n} \to +\infty$ the limit is $-\infty$ (so for the original quotient we get the limit $0$), and for $\alpha < 1$ (just for completeness) the term $(\log n)(\log \log n)^{1-\alpha}$ in the subtrahend grows faster than $\log n$, so we again have the limit $-\infty$ for the logarithm of the quotient. Note that the values of $A$ and $B$ are completely irrelevant.
