I tried using this inequality: $\tau (n) < (\frac{2}{\log 2} \log n)^{2^{x}}n^{1/x}$ which gives : $\frac{\log \tau (n)}{\log n} < \frac{(2^{x})\log(2)\log\log n -x\log n}{(\log \log 2) \log n} $
however then the limit will approach infinity so it isnt useful.
According to the text, $\lim_{n\to\infty}\frac{\log \tau (n)}{\log n}$ is supposed to be 0. How does one see that? (tau is the number of divisor function, x \in \mathbb{R}_{>0})