Evaluating the limit $\lim_{x\to\infty}\frac{(1+1/x)^{x^2}}{e^x}$ 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.

Compute the limit of the logarithm of your function:
$$
\lim_{x\to\infty}\log\frac{(1+1/x)^{x^2}}{e^x}=
\lim_{x\to\infty}(x^2\log(1+1/x)-x)
$$
Now set $x=1/t$:
$$
\lim_{x\to\infty}(x^2\log(1+1/x)-x)=
\lim_{t\to0^+}\frac{\log(1+t)-t}{t^2}
$$

 The limit becomes $\displaystyle\lim_{t\to0^+}\frac{t-t^2/2+o(t^2)-t}{t^2}=-\frac{1}{2}$. So your original limit is $e^{-1/2}$.

