Find the limit
lim
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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