# How to show $\lim_{n\to \infty}\sqrt{n}^n (1 - (1 - 1/(\sqrt{n}^n))^{2^n})/2^n = 1$?

How can you show the following?

$$\lim_{n\to \infty}\frac{\sqrt{n}^n \left(1 - \left(1 - \frac{1}{\sqrt{n}^n}\right)^{2^n} \right)}{2^n} = 1$$

It certainly seems to be true numerically when I plot it.

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In the nummerator, you can use $(1-\frac{1}{\sqrt{n}^n})^{2^n} \approx 1-\frac{2^n}{{\sqrt{n}^n}}$ when $n$ is large. – user137846 Mar 27 '14 at 12:32
Both numerator and denominator go to $\infty$. So it's not a bad idea to say $\lim \frac{a}{b} = \lim \frac{\ln a} {\ln b}$. That'll let you cancel an $n$ on top and bottom. My guess is that from there, you probably just need algebra and L'hopital's rule. – John Hughes Mar 27 '14 at 12:42
John, that assertion is false. Consider $f(n) = n^2$ and $g(n) = n$. Then $f(n)/g(n) = n \rightarrow \infty$ as $n \rightarrow \infty$. However, $\log(f(n))/\log(g(n)) = 2\; \forall n$. Clearly, the limits are not equal. – Chris K Mar 27 '14 at 13:19

Let $x=\sqrt n^n$, then $2^n=x^2\left(\frac2n\right)^n$. Then we have \displaystyle\begin{align}\lim_{n\to\infty}\frac{x\left(1-\left(1-\frac1x\right)^{x^2\left(\frac2n\right)^n}\right)}{x^2\left(\frac2n\right)^n}=&\lim_{n\to\infty}\frac{1-e^{-x\left(\frac2n\right)^n}}{x\left(\frac2n\right)^n} =\lim_{n\to\infty}\frac{1-e^{-y}}{y}\end{align} where $y=2^nn^{-\frac n2}\to0$ as $n\to\infty$. Hence the limit is 1.
Of course this hinges on the quite nonrigorous and quite unjustified step replacing $(1-u)^v$ by $e^{-uv}$ for some varying $u$ and $v$. – Did Mar 27 '14 at 13:07
@Did It's neither of $1^\infty$ nor of $\infty^0$ form so I think it rigorous. – Shuchang Mar 27 '14 at 13:30
Sorry but this is very much of the $1^\infty$ form, more precisely of the form $\infty(1-(1-0)^\infty)$. – Did Mar 27 '14 at 15:57
@octonots The last step is directly followed by either expansion of $e^y$ or l'Hopital's rule. I will see that post. – Shuchang Mar 28 '14 at 0:52