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In one of the exercises I got, I was required in order to proceed to calculate the limit:

$\lim_{n\to \infty} (\frac 1 {e^n} (1+ \frac 1 n)^{n^2})$

I checked in the solution sheet to see if the answer will give me a clue, and the answer is supposed to be $\frac 1 {\sqrt e}$ but I still can't see how I get there... Can anyone please give me a clue? :)

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$\frac 1 {e^x} (1+ \frac 1 x)^{x^2}=\left(\frac{\left(1+\frac{1}{x}\right)^x}{e}\right)^x=e^{x\left(ln\left(1+\frac{1}{x}\right)^x-lne\right)}=e^{\frac{ln\left(1+\frac{1}{x}\right)^x-1}{\frac{1}{x}}}$

Now apply L'Hôpital's rule twice on the exponent and you will get the result.

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  • $\begingroup$ Now that solved it! Thanks!! $\endgroup$
    – AsafHaas
    Mar 5 '17 at 17:26
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If $a_n = \frac 1 {e^n} (1+ \frac 1 n)^{n^2} $, then

$\begin{array}\\ \ln a_n &=n^2\ln (1+ \frac 1 n)-n\\ &=n^2(\frac1{n}-\frac1{2n^2}+O(\frac1{n^3}))-n\\ &=(n-\frac12+O(\frac1{n}))-n\\ &=-\frac12+O(\frac1{n})\\ &\to -\frac12\\ \end{array} $

so $a_n \to e^{-1/2}$.

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