A Tricky Limit: $(1 - \frac{c}{n}\log n )^{1-n}$ I'm stuck with this limit $(1 - \frac{c}{n}\log n )^{1-n}$ as $n \rightarrow \infty$ where $c < 1$. I tried to plot the limit and it looks like it goes to infinity, although very slowly, but I can't prove it. Any ideas?
 A: Near $x=0$, $\log(1+x)=x+O(x^2)$ so as $n\to\infty$, we get that
$$
\begin{align}
(1-n)\log(1-\frac{c}{n}\log(n))
&=(1-n)\left(-\frac{c}{n}\log(n)+O\left(\left(\frac{\log(n)}{n}\right)^2\right)\right)\\
&=\frac{c(n-1)}{n}\log(n)+O\left(\frac{\log(n)}{n}\log(n)\right)\\
&\to\infty\text{ (like }c\log(n)\text{)}
\end{align}
$$
if $c>0$. Thus, $(1 - \frac{c}{n}\log n )^{1-n}\to\infty$ like $n^c$.
A: You can factor out $(1 - {c \over n}\log(n))$ which converges to $1$ and thus you are looking for
$$\lim_{n \rightarrow \infty} (1 - {c \over n}\log(n))^{-n}$$
$$= \lim_{n \rightarrow \infty} \bigg((1 - {c \over n}\log(n))^{{n \over c \log n}}\bigg)^{-c\log n}$$
Since ${\displaystyle \lim_{\epsilon \rightarrow 0} (1 - r)^{1 \over r} = {1 \over e}}$, the expression inside the large parentheses goes to ${1 \over e}$ as $n$ goes to infinity. Since $({1 \over e})^{-c\log n} = n^c$, this means the expression diverges to infinity as $n^c$ does. (Well faster than $n^{c'}$ for any $c' < c$.)
A: Some graph samples shows that as x gets larger the limit goes larger.

Edit:
I removed my steps due to corrections suggested below.
