Calculate $\liminf_{n \to \infty} d(n)/\log(n)$ and $\limsup_{n \to \infty} d(n)/\log(n)$ Calculate $\liminf_{n \to \infty} d(n)/\log(n)$  and $\limsup_{n \to \infty} d(n)/\log(n)$
$d(n)$ being the number of positive divisors for a positive integer $n$.
So from what I understand of limit superior and inferior, here's how I would approach the problem.
The lower bound of $d(n)$ as n approaches infinity is 2, as there infinitly many prime numbers, and the only positive dividers of prime numbers are 1 and itself.
The upper bound of $d(n)$ is infinity, as if $n$ is composite, then it would it have infinite amount of positive disviors as n approaches infinty.
So, as such, the  $\liminf_{n \to \infty} d(n)/\log(n)$ would be $2/\infty$, which would just $0$,
while $\limsup_{n \to \infty} d(n)/\log(n)$ would be $\infty/\infty$, would would just be $1$.
Or did get them reversed/wrong?
 A: First, for all prime number $p$, $d(p)=2$ and thus $\liminf\limits_{n\rightarrow +\infty}\frac{d(n)}{n}=0$. As for the upper bound, let $n_k:=p_1\ldots p_k$, then $d(n_k)=2^k$ and
$$ \ln n_k=\vartheta(p_k)\leqslant\pi(n_k)\ln p_k=k\ln p_k $$
Thus $d(n_k)\geqslant 2^{\frac{\ln n_k}{\ln p_k}}$. But since $\ln n_k=\vartheta(p_k)\gg p_k$, we have $d(n_k)\geqslant\exp\left(\frac{\ln (2)\ln(n_k)}{\ln\ln n_k+\ln A}\right)$ for some constant $A>0$ such that $p_k\leqslant A\ln n_k$ for all $k\geqslant 1$. We thus have $\lim\limits_{k\rightarrow +\infty}\frac{d(n_k)}{\ln n_k}=+\infty$ and finally $\limsup\limits_{n\rightarrow +\infty}\frac{d(n)}{\ln n}=+\infty$.
A: You are right that the inferior limit is $0$.
For the upper limit you have to identify "small" integers with lots of divisors. An extreme case are the powers of $2$: $2^k$ has the $k$ divisors $1, 2, \dotsc, 2^k$, for $k + 1$ in all. Would need to prove no number up to $2^k$ has more than $k + 1$ divisors (or find numbers with more) to derive your limit.
