Let $\omega(n)$ be the number of distinct prime divisors of $n$, and let $\tau(n)$ be the number of divisors of $n$. Then, one can show that $ \tau(n) \le \exp \left( \frac{ \log 2 \log n}{\log \log n} \left( 1 + O\left( \frac{\log \log \log n}{\log \log n} \right) \right) \right)$ (see, for instance, G. Tenenbaum's Introduction to Analytic and Probabilistic Number Theory, chapter 5).
Then, since $2^\omega(n) \le \tau(n)$, we may conclude that $ \omega(n) \le (1+ o(1)) \frac{\log n}{\log \log n}$ as $n \rightarrow \infty$. By directly working with numbers which are the product of the first $k$ primes, a lower bound for $\tau$ is attainable which yields a lower bound for $\omega$ of the same order. In short, one has $\limsup_{n \rightarrow \infty} \frac{\omega(n)}{(\log n)/(\log \log n)} = 1.$