Finding a lower bound for the expression $\ln^k(n)$ Problem:
Assume that $k \geq 1$ and $\epsilon \geq 0$, is $\ln^k(n) \in$ $\Omega( n^\epsilon)$?
Answer:
In the special case of $\epsilon = 0$, I claim the answer is no. I suspect the person who wrote this
problem meant to write: assuming $\epsilon &gt; 0$.
I claim that if we differentiate $\log^k(n)$, k times then we get an expression with $n^{-k}$ in it. If
we differentiate $n^{e}$  k times then we get an expression with $n^{\epsilon- 2}$ in it. Let me illustrate
this with an example. Suppose I pick $k = 2$. Then if I differentiate the first expression twice, I get:
$$ \frac{ 2 - 2 \ln(x) }{x^2} $$
When I differentiate the second expression twice, I get:
$$ \epsilon ( \epsilon - 1 ) n ^ { \epsilon - 2 } $$
Hence, I claim the answer is no. Do I have this right?
The notation $\log^k n$ most likely stands for $(\ln n)^k$ (this is a common abuse of notation).
In this case the claim is trivially when $\varepsilon=0$ true since $n^\varepsilon=1$ and $\ln^k n = \Omega(1)$.
For constant $\varepsilon&gt;0$ and $k \ge 1$ you have:
$$
\lim_{n \to +\infty} \frac{\ln^k n}{n^\varepsilon} =
\lim_{n \to +\infty} \left( \frac{\ln n}{n^\frac{\varepsilon}{k}} \right)^k = 
\left( \lim_{n \to +\infty}  \frac{\ln n}{n^\frac{\varepsilon}{k}} \right)^k =
\left( \lim_{n \to +\infty} \frac{1}{n \cdot \frac{\varepsilon}{k} \cdot n^{\frac{\varepsilon}{k}-1}}\right)^k
= \left( \lim_{n \to +\infty} \frac{k}{\varepsilon n^\frac{\varepsilon}{k}} \right)^k
 = 0,
$$
showing that $\ln^k n = o(n^\varepsilon)$, and hence that the claim is false.