Determine the asymptotic behavior of $f(n)$ in relation to $g(n)$

Question

1. $f(n)=n^\sqrt{n}, g(n)=2^n$
2. $f(n)=10^{\log\log n}, g(n)=\log n$

Note: $\log$ is in base 2.

For section #1, I tried to evaluate the limit $\lim_{n\to\infty} \frac{2^n}{n^\sqrt{n}}$ but got stuck along the way.

For section #2, this is what I thought about:

Since $\log$ is a monotone function we can evaluate $f'(n)=\log(10^{\log\log n})$ and $g'(n)=\log\log n$. But, $f'(n) = \log\log n \log 10$.

So we have that $f(n) = \Theta(g(n)$

I'd be glad to get help with section #1 and get solution validation for section #2.

Thanks in advance!

Answer 1

Take logarithms

1. $\log_2 f(n) = \sqrt{n} \log_2 n$ while $\log_2 g(n) = n $. Note that $\log_2 n$ grows more slowly than $\sqrt{n}$

2. $\log_2 f(n) = (\log_2 10)(\log_2\log_2 n)$ while $\log_2 g(n) = \log_2\log_2 n $. So $\log_2 f(n)$ is more than three times $\log_2 g(n)$ and so $f(n)$ is more than the cube of $g(n)$

Answer 2

Hint:
$$\frac{2^n}{n^\sqrt{n}}=\frac{2^{n}}{2^{\sqrt{n}\log_2{n}}}=2^{n-\sqrt{n}\log_2{n}}=2^{\sqrt{n}(\sqrt{n}-\log_2{n})}.$$