# Trouble with Big-O notation proof by definition

Let $$a,b>0$$. Prove $$\left(\log\left(n\right)\right)^{a}=O\left(n^{b}\right)$$.

I'm supposed to find an algorithm to find the log(n) largest elements in an array and return them sorted and explain why it's bounded by $$O(n^{b})$$, but the thing is I'm having trouble proving the above and I'm kind of stuck.. Any tips?

As $$a$$ is fixed, then you can omit it from requirements because for $$a \gt 0$$ inequality $$(\log n)^a \leqslant C n^b$$ is equivalent $$\log n \leqslant C^{\frac{1}{a}}n^{\frac{b}{a}}$$. Denoting $$0 \lt \frac{b}{a}=\alpha$$ we obtain, that it's enough to proof $$\log n \in O(n^\alpha)$$ for $$\alpha \gt 0$$.
To simplify, we can consider $$\log n$$ with base $$e$$, because the difference with the other base will be constant factor.
So, last step will be $$\lim\limits_{n \to \infty}\frac{\ln n}{n^\alpha} = \lim\limits_{n \to \infty}\frac{1}{\alpha n^\alpha}=0$$
We see, that a more strong result with $$o$$-little is fair: $$\forall \alpha \gt 0$$ we have $$\log n \in o(n^\alpha) \subset O(n^\alpha)$$.