# limit of $(1-k/n)^{k}$ where $k$ depends on $n$

Define $$L(k) = \lim_{n \rightarrow \infty} (1-\frac{k}{n})^{k}$$, where $$1 \leq k \leq n$$.

For "low" values of $$k$$ (e.g. $$k = c_0$$ independent of $$n$$) it holds that $$L = 1$$, and for "high" values of $$k$$ (e.g. $$k = n - 1$$) it holds that $$L = 0$$. I am trying to compute the "phase transition range", i.e. the range $$O = [1,f(n)]$$ such that for all $$k \in O$$, $$L(k) = 1$$ and the range $$Z = [g(n), n]$$ such that for all $$k \in Z$$, $$L(k) = 0$$ (where $$f,g$$ are as tight as possible).

Any ideas?

• Take logarithms ? Apr 5, 2020 at 14:48
• Tried that. But I couldn't see how $k \log(1-\frac{k}{n})$ is better than the original expression (it helps to see the behavior for a given $k$, but I couldn't derive the range from it) Apr 5, 2020 at 14:56
• If k is much smaller than n then $\log(1-k/n)\approx -k/n$ Apr 5, 2020 at 14:58
• Indeed, but this would only help to understand what happens when $k$ is "small". It won't help to understand what happens when $k = \Theta(n)$, and I don't see how it helps to determine the range Apr 5, 2020 at 15:05
• If $k=m\sqrt n$ the log is roughly $-m^2$, $L\approx \exp(-m^2)$ and it is in this interval that $L$ goes from $1$ to $0$ Apr 5, 2020 at 15:10

If $$k= m\sqrt n$$ as $$n$$ gets large, then $$k/n$$ gets small, so $$\ln L(k)=k\ln(1-k/n)\approx k(-k/n) =-m^2\\L(k)\approx e^{-m^2}$$