Finding limit of sequence $a_n=\frac{\alpha(n)}{n}$ Given sequence $a_n=\frac{\alpha(n)}{n}$, where $\alpha(n)$=total no. of primes by which $n$ is divisible, find $\lim_{n\to \infty}a_n$. 
Example: $\alpha (8)=1$, as $2$ is the only prime by which $8$ is divisible.
Observing that $\alpha(n)\lt n$ for all $n\in \mathbb N$, the given sequence is bounded. Then I tried using the fact that every $n$ has prime factorisation.
But I don't know how to proceed further. Any hints?
 A: Hint: You can have a stricter bound than just $n$ for $\alpha(n)$. If every prime factor were a $2$ and repeated prime factors were allowed, then $\alpha(n) = \log_2(n)$, which is an upper bound because other prime factors have to be larger than $2$. Then, an upper bound on the limit is $$\lim_{n \to \infty}\frac{\log_2(n)}{n}$$
A: I am posting an answer here. I would like to thank @Teresa Lisbon who's helped me a lot to solve this exercise.
Since every natural no. $n$ has a prime factorization, suppose that $n=n_1^{p_1}n_2^{p_2}\cdots n_k^{p_k}$ for prime nos. $n_i$'s where $i\in\{1,2,3,\cdots,k\}$. It follows that $\phi(n)=k$ as for every $i\in\{1,2,3,\cdots,k\}$, we have prime no. $n_i$ such that $n_i|n$. 
$\begin{align}
\log_2n &=p_1\log_2n_1+p_2\log_2n_2+\cdots+p_k\log_2n_k\\
        &\ge p_1\log_22+p_2\log_22+\cdots+p_k\log_22\\
        &=p_1+p_2+\cdots+p_k \ge 1+1+\cdots+1 \text{  ($k$ times)}=k\\
\implies &\log_2n\ge k= \phi(n)\\
\implies &0\le\frac{\phi(n)}{n}\le\frac{\log_2n}{n}\end{align}$ $\tag 1$ 
Since $n^{\frac 1{n}}\to 1$, by continuity of $\log_2n$ we have $\frac{\log_2n}{n}=\log_2(n^{\frac 1{n}})\to \log_2(1)=0$
By squeeze theorem in $(1)$, it follows that $\frac{\phi(n)}{n}\to 0$ 
