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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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    Oops.. I messed up the expression by missing $\log$. Read $L:=\lim_{n \to \infty} \exp{(1-n) \log {(\frac{(n-c \log n)}{n}}}$2011-10-10

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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.

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

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    I tried to address the comment to 2 people but I was not allowed to by the $f$orum.2011-10-10