Finding $f$ satisfies $\limsup_{n\to\infty}\frac{\tau(n)}{f(n)}=1$

Question

Is there a known result on finding the function $f$ satisfies

$$\limsup_{n\to\infty}\frac{\tau(n)}{f(n)}=1?$$ where $\tau(n)$ is the number of factor(s) of $n$.

Related:Prove that $d(n)\leq 2\sqrt{n}$
It shows that $f(n)=O(\sqrt n)$.
Some Ideas
Also, letting $n=2^m,m\in\mathbb Z$ gives $f(n)=\Omega(m)$, or $f(n)=\Omega(\ln n)$.
Factorize $n$ into $\prod_ip_i^{\alpha_i}$ gives $\tau(n)=\prod_i(\alpha_i+1)$. So probably the next step to do is to estimate $p_i$.
Wikipedia gives $p_n\le n\ln n+n\ln\ln n$ when $n\ge 6$. Can we use this result to give the better bounds for $f$?

Answer 1

The best result in this direction is due to S. Wigert (1907):$$\limsup_{n\to\infty}\frac{\log\tau (n) \log\log(n)}{\log(2)\log(n)}=1.$$ It comes as a corollary to certain estimates of $\max_{n\le x} \tau(n)$, the strongest of which is due to S. Ramanujan (1915).