# 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?

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Consider taking the limit of $(1-n)\log(1-(c\log n)/n)$... –  Guess who it is. Oct 10 '11 at 14:39
Wolfram Alpha says, $\lim_{n\to\infty}(1-\frac cn\log n)^{1-n} = \infty$... –  FUZxxl Oct 10 '11 at 14:41
Hence the sequence is $n^{c+o(1)}$, which is slow but not so slow... –  Did Oct 10 '11 at 14:42
For $0<c<1$ the my result came $\infty$. –  gaurav Oct 10 '11 at 14:43
I tried writting the expression as : $L:=\lim_{n \to \infty} \exp{(1-n+\frac{(n-c)}{n} \log n)}$. –  gaurav Oct 10 '11 at 14:53

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 $n^c$.

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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 $c' < c$.)

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Some graph samples shows that as x gets larger the limit goes larger.

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

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As n grows larger and larger, (c/n)log(n) goes to c... Hmmmm, well, it does not. –  Did Oct 10 '11 at 15:13
Well, it tends to $0$. And for $c \in (0,1)$, one might say that this limit tends to c. –  gaurav Oct 10 '11 at 15:19
Thanks to @DidierPiau and gaurav for help –  Emmad Kareem Oct 10 '11 at 19:26
@gaurav, ???  –  Did Oct 10 '11 at 19:43
I tried to address the comment to 2 people but I was not allowed to by the forum. –  Emmad Kareem Oct 10 '11 at 19:44