Show that $\log^{i} n \in O(n^{j})$ for $i,j > 0 $

Question

I want to show that

$$\log^{i} n \in O(n^{j})$$

I tried to apply L'Hospital and came up with the following:

$$\lim\limits_{n \rightarrow \infty}{\frac{\log^{i} n}{n^{j}}} =$$ $$\lim\limits_{n \rightarrow \infty}{\frac{\log^{i-1} (1/n) n}{j*n^{j-1}}} =$$ $$\lim\limits_{n \rightarrow \infty}{\frac{\log^{i-1} n}{n^{j}}}$$

Now I am kind of stuck since I don't no how to make a statement out of this. Help would be greatly appreciated!

Answer 1

Hint: The power $j$ on $n^j$ in the denominator stays the same, but your power on the logarithm drops by $1$. What happens if you do this over and over so that the power on $\log$ becomes less than or equal to zero? Then you can put it in the denominator by switching signs on the power and hopefully be able to determine the limit from there, assuming $j > 0$.