Solving a Limit with L'Hospital's Rule with 2 Unknown Constants I am trying to find this limit:
$$ \lim_{n \to \infty} \frac{(\log_{2}n)^k}{n^p} $$ where $k$ and $p$ are constants and $k \geq 1$, $p \gt 0$. 
The limit equates to $0$, but I need someone to help explain this derivation. Thank you.
 A: It is easy to prove the standard result $$\lim_{x \to \infty}\frac{\log x}{x^{a}} = 0\tag{1}$$ where $a > 0$. To establish this we need to choose any number $b$ such that $0 < b < a$. Then we have $$b\log x = \log x^{b} < x^{b} - 1 < x^{b}$$ for $x > 1$ (this is standard inequality of $\log x$ namely $\log x < x - 1$ for $x > 1$). Thus we have $$\log x < \frac{x^{b}}{b}$$ and therefore if $x > 1$ then we have $$0 < \frac{\log x}{x^{a}} < \frac{1}{bx^{a - b}}$$ Applying Squeeze theorem when $x \to \infty$ and noting that $(a - b) > 0$ we see that $(\log x)/x^{a} \to 0$ as $x \to \infty$.
Now we come to question posted here. We have
\begin{align}
L &= \lim_{n \to \infty}\frac{(\log_{2}n)^{k}}{n^{p}}\notag\\
&= \lim_{n \to \infty}\frac{(\log n)^{k}}{n^{p}(\log 2)^{k}}\notag\\
&= \frac{1}{(\log 2)^{k}}\lim_{n \to \infty}\left(\frac{\log n}{n^{p/k}}\right)^{k}\notag\\
&= \frac{1}{(\log 2)^{k}}\cdot(0)^{k} = 0\notag
\end{align}
A: Rewrite the limit as
$$
\lim_{n\to\infty}\left(\frac{\log_2n}{n^{p/k}}\right)^{\!k}
$$
and compute the inner limit:
$$
\lim_{n\to\infty}\frac{\log_2n}{n^{p/k}}=
\lim_{n\to\infty}\frac{\log_2e}{(p/k)n^{p/k}}=…
$$
The derivative of $x\mapsto \log_2x$ is $\frac{\log_2e}{x}$, so applying l'Hôpital gives the above expression.
A: Another way: write $\log_2 t = \frac{\ln t }{\ln 2}$ and set $\ln n  = t$, s you get $\frac{t^k}{e^{pt}}$. 
