Prove that $x^\frac{1}{3} = o(2^{\sqrt{log_2 x}}) $ and $ 2^{\sqrt{log_2 x}} =o(10^x)$ for $ x \rightarrow \infty$

Question

I'm trying to prove the asymptotic hierarchy above. It makes intuitive sense, since $2^{\sqrt{log_2 x}}$ is somewhat slower than linear but still faster than $x^q, 0<q<1$. But I don't know how to give a rigorous proof. I tried l'hopital, but that didn't seem to give handier expressions. Also tried some kind of substitution, but to no avail. How do I handle this $2^{\sqrt{log_2 x}}$?

Answer 1

If you use the identity: $x = 2^{log_2x}$

$x^{\frac{1}{3}}=2^{log_2(x^{\frac{1}{3}})}$
$\quad \;\; =2^{\frac{1}{3}log_2x}$
$\quad \;\; =2^{\sqrt{{\frac{1}{9}log_2x\;log_2x}}}$
$\quad \;\; \ge 2^{\sqrt{{\frac{1}{9}9\;log_2x}}}\quad\quad$ for $x \gt 2^9$

$\quad \;\; = 2^{\sqrt{log_2x}}\quad\quad\;\quad$ for $x \gt 2^9$

But this shows that $x^{\frac{1}{3}} \ne o(2^{\sqrt{log_2x}})$

Did you mean 'little-omega' instead of 'little-o'?

For the second part:

$10^x = 2^{log_210^x}$
$\quad\;\;\;= 2^{xlog_210}$
$\quad\;\;\;> 2^{x}$
$\quad\;\;\;> 2^{log_2x}$
$\quad\;\;\;> 2^{\sqrt{log_2x}}\quad\quad\quad$ for $x \gt 2$