Find the limit
$$
\lim_{x \to \infty} \frac{e^{x}}{(1+\frac{1}{x})^{x^{2}}}.
$$
I know that the limit is of indeterminate form type $\frac{\infty}{\infty}$, but it seems using L'Hopital's rule directly does not help here. Do I somehow use the fact that $\lim_{x \to \infty} (1+\frac{1}{x})^{x}=e$?
Answer
For any $x>1$,
$$ x^2\log\left(1+\frac{1}{x}\right) = x -\frac{1}{2}+O\left(\frac{1}{x}\right),$$
so:
$$ x-x^2\log\left(1+\frac{1}{x}\right) = \frac{1}{2}+O\left(\frac{1}{x}\right). $$
By exponentiating such identity, you get that the limit is $e^{1/2}=\color{red}{\sqrt{e}}$.
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