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I’m trying to find for what values for $p$ will cause $$\displaystyle\lim_{n\to\infty} \frac{\ln^p{n}}{n} = 0$$ I believe that one criteria for a limit approaching zero is that the top should be going to infinity slower than the bottom; however in this expression, it seems that I can make the top grows as fast as I can by adjusting $p$.

I tried to plot the graph of

$$

f(p) = \displaystyle\lim_{n\to\infty} \frac{\ln^p{n}}{n}

$$

and indeed, $\forall p$ yields $0$.

I was thinking if I can prove it without plotting every value. This is what I did:

- Using integral definition to solve this integral
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$$

\begin{align}

&\displaystyle\lim_{n\to\infty} \frac{\ln^p{n}}{n} \\

\overset{L}{=}& \displaystyle\lim_{n\to\infty} \frac{p\ln{(n)}^{p-1}}{n} \\

\end{align}

$$

But then I got stuck here. It seems like this is not the way of solving it. But if this is not, then what is the way of solving this?

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For $p>0$ and $q>0$, we have that

$$

\lim_{x\to\infty}\frac{\log^px}{x^q}

=\lim_{x\to\infty}\Bigl(\frac{\log x}{x^{q/p}}\Bigr)^p

=\Bigl(\lim_{x\to\infty}\frac{\log x}{x^{q/p}}\Bigr)^p

=\Bigl(\lim_{x\to\infty}\frac1{(q/p)x^{q/p}}\Bigr)^p=0

$$

using continuity and l’Hôpital’s rule.

**Hint** What happens if you repeat the L’Hôpital’s rule $p$ times?

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