Limit in number theory

It is actually true that $\lim_{n\to\infty} \frac{d(n)}{n^c}=0$ for any $c>0$, but if you only want it for $c=1$, the simplest proof would be the following. The divisors $\delta\mid n$ can be organized into pairs $(\delta,n/\delta)$, and in every pair the smallest divisor is $\min\{\delta,n/\delta\}\le\sqrt n$. It follows that the number of pairs is at most $\sqrt n$. Thus $d(n)\le 2\sqrt n$ and the assertion follows.

Hint: For every divisior $p$ greater than $\sqrt{n}$ of $n$, there is a divisor $q$ smaller than $\sqrt{n}$ such that $n=pq$. It follows that we can divide the divisiors in pairs where one of the elements is smaller than $\sqrt{n}$, and hence $d(n) \leq 2\sqrt{n}$. Can you use this to evaluate the limit?