TITLE: Evaluating the limit $\lim_{x\to\infty}\frac{(1+1/x)^{x^2}}{e^x}$
QUESTION [8 upvotes]: Could anybody show me step by step why the following equality holds?
$$\lim_{x\to\infty}\frac{(1+1/x)^{x^2}}{e^x} = e^{-1/2}$$
The most obvious method gives you 1 as an answer, but I understand that only the limit of $(1+1/x)^x$ is $e$, but expression is not actually equal to $e$. And now I am stuck.

REPLY [4 votes]: Say $$y = \frac{\left( 1 + \frac{1}{x}\right)^{x^2}}{e^x}$$ Then $\log(y) =x^2 \log\left( 1 + \frac{1}{x}\right)-x$. We change variables $t=\frac{1}{x}$. Then $\log(y)=\frac{1}{t^2} \log(1+t) - \frac{1}{t}$.
Near $t=0$, $$\log(1+t) = t - \frac{t^2}{2}+\frac{t^3}{3}- \frac{t^4}{4} +\ldots$$
So near $t=0$, $$\log(y) = \left(\frac{1}{t} - \frac{1}{2} + O(t)\right) - \frac{1}{t} = - \frac{1}{2} + O(t) $$
Going back in terms of $x$, for large $x$
$$\log(y) = - \frac{1}{2} + O\left(\frac{1}{x}\right) $$
Hence $$\lim_{x\rightarrow\infty} \log(y) = -\frac{1}{2}$$
exponentiating we have the result.