# Find $f\colon \Bbb{N}\to\Bbb{N}$ such that $f = n^{o(1)}$ and for every $c\in\Bbb{N}$, $f(n)=\omega(\log^c n)$

Find $f\colon \Bbb{N}\to\Bbb{N}$ such that $f = n^{o(1)}$ and for every $c\in\Bbb{N}$, $f(n)=\omega(\log^c n)$.

It looks like I need to find a very small growing function $g$ that will satisfy $$\lim_{n\to{\infty}} \frac{1}{g} = 0$$ and also $$\lim_{n\to{\infty}} \frac{\log^cn}{n^{\frac{1}{g}}} = 0$$

My first attempt was $g(n) = \log n$ but according to wolfram alpha, the second limit was not satisfied, I then tried $g(n) = \log\log n$ which according to wolfram does the trick.

But I'm not sure how to prove that $$\lim_{n\to{\infty}} \frac{\log^cn}{n^{\frac{1}{\log\log n}}} = 0$$

Im looking for a way to prove this, or find a simpler solution than $f=n^{1/\log\log n}$.

Take the log of your ratio to get $$c\log\log n - \frac{\log n}{\log\log n} = \frac{c\log^2 m - m}{\log m}, \quad \text{where } m = \log n.$$ It is not hard to show that this tends to $-\infty$ as $m\to\infty$, and so as $n\to\infty$, which means that your ratio tends to zero as $n\to\infty$.
Indeed, since $c\log^2 m = o(m)$, for large enough $m$ we have $c \log^2 m - m \leq -m/2$. Since $2\log m = o(m)$, it follows that $\frac{m}{2\log m} \to \infty$ and so $\frac{-m}{2\log m} \to -\infty$.