real analysis - Show $forallvarepsilon>0,lim_{ntoinfty}frac{#{text{positive divisors of n}}}{n^varepsilon}=0$

Show that $\forall\varepsilon>0,$

$$\lim_{n\to\infty}\frac{\#\{\text{positive divisors of n}\}}{n^\varepsilon}=0$$

I'm trying to solve this problem for a long time, but I'm really stuck I have totally no idea where to start.
I tried replacing $\varepsilon$ by $\frac{1}{k}$ where $k$ is a natural number, and show the statement is true for all $k$ by induction, but I haven't succeeded and it doesn't seem promising.

If you give me any advice or comment, I would greatly appreciate.

Thank you.

Answer

The solution for this problem falls out from the following fact.

\begin{equation}
e^{O(\frac{log(n)}{loglog(n)})}=d(n)
\end{equation}

So,

\begin{equation}
\lim_{n->\infty}\frac{d(n)}{n^\epsilon}=\frac{e^{\frac{Clog(n)}{loglog(n)}}}{n^\epsilon}
\end{equation}
For some constant C.

Therefore,
\begin{equation}

\lim_{n->\infty}\frac{e^{\frac{Clog(n)}{loglog(n)}}}{n^\epsilon}=\lim_{n->\infty}\frac{n^\frac{C}{loglog(n)}}{n^\epsilon}=\lim_{n->\infty}n^{\frac{C}{loglog(n)}-\epsilon}
\end{equation}

You can add a nice proof regarding epsilon and deltas to show from here that

\begin{equation}
\lim_{n->\infty}\frac{C}{loglog(n)}-\epsilon=-\epsilon\\and\\
\lim_{n->\infty}n^{-\epsilon}=0\\
implies\\
\lim_{n->\infty}n^{\frac{C}{loglog(n)}-\epsilon}=0

\end{equation}

This completes our proof. Many websites have the first point written and I believe you may be able to find out more about it from other posts.