I just came across the statement that any power of $\log(x)$ grows much slower than any positive power of $x$, i.e. $(\log(x))^a$ = $o(x^b)$ for arbitrary $a$ and $b >0$. How can I prove it?
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2Related is the recent question Why does $\log^{100}(n)$ grow asymptotically slower than $n^{7/8}$. Note that the answer by Bernard is exactly what you are asking. – Dave L. Renfro Jan 30 '21 at 16:45
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Oh! thanks a lot! – 김세현 Jan 31 '21 at 06:57