The limit of $n^2 \log^n(1 - \frac{c \log n}{n})$ Maple tells me that $\lim_{n \to \infty} n^2 \log^n(1 - \frac{c \log n}{n}) = 0$ for any constant $c$, but I can't find a way to prove it. Any suggestions?
 A: Well
$$\log\left(1-c\frac{\log n}{n}\right)=-c\frac{\log n}{n}
+O\left(\frac{(\log n)^2}{n^2}\right)$$
and so
$$n^2\log\left(1-c\frac{\log n}{n}\right)^n=(-c)^n\frac{(\log n)^n}{n^{n-2}}
\left(1+O\left(\frac{\log n}{n}\right)\right)^n.$$
Now
$$n\log\left(1+O\left(\frac{\log n}{n}\right)\right)=O(\log n)$$
so that
$$\left(1+O\left(\frac{\log n}{n}\right)\right)^n$$
grows as at most a polynomial in $n$. The $n^n$ in the denominator
will swamp everything else...
A: The Taylor series expansion $- \log (1-x) = \sum_{k \ge 1} \frac{x^k}{k}$ shows that $\log(1 - x) = O(x)$ for small $x$.  So $\log \left( 1 - \frac{c \log n}{n} \right) = O \left( \frac{\log n}{n} \right)$, hence the absolute value of the expression in question is bounded by a constant times $\frac{(\log n)^n}{n^{n-2}}$ which rather clearly goes to zero.  
This is a good example of why it's more flexible to use big-O notation than to deal directly with the definition of a limit.
A: The case $c=0$ is clear, and then for the rest of its values it's enough to consider and calculate $\lim_{n\rightarrow\infty} n^2 \log^n(1 + \frac{|c| \log n}{n})$. Then we have that: 
$$\lim_{n\rightarrow\infty} n^2 \log^n(1 + \frac{|c| \log n}{n})=\lim_{n\rightarrow\infty} n^2 \left(\frac{|c|\log n}{n}\right)^{n} \left(\left(\frac{\log(1 + \frac{|c| \log n}{n})}{(\frac{|c| \log n}{n})}\right)^{\frac{n}{|c|\log n}}\right)^{|c|\log n}= \lim_{n\rightarrow\infty} n^{2} \left(\frac{|c|\log n}{n}\right)^{n} \frac{1}{e^{\frac{|c|\log n}{2}}}=0.$$
The auxiliary limit i resorted to is $\lim_{x\to0} \left({\frac{\ln(1+x)} {x}}\right)^\frac{1}{x}= \frac{1}{\sqrt{e}}.$  
The proof is complete.
