calculus - What is? $lim_{x to infty} left(frac{e}{left( 1 + frac {1}{x} right)^x} right)^x$

Find $$\lim_{x \to \infty} \left( \dfrac{e}{\left( \left( 1 + \frac {1}{x} \right)^x \right)} \right)^x.$$

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$\lim_{x \to \infty} \left( \dfrac{e}{\left( \left( 1 + \frac {1}{x} \right)^x \right)} \right)^x = \lim_{x \to \infty} \left( \dfrac {e}{e} \right)^x = \lim_{x \to \infty} 1^x = \infty$

Why is this incorrect?

Answer

Take the logarithm,

$$\begin{align} \log \left(\frac{e}{\left(1+\frac1x\right)^x}\right)^x &= x\log \frac{e}{\left(1+\frac1x\right)^x}\\ &= x\left(1 - x \log \left(1+\frac1x\right)\right)\\ &= x\left(1 - x\left(\frac1x - \frac{1}{2x^2} + O(x^{-3})\right) \right)\\ &= x\left(1 - 1 + \frac{1}{2x} + O(x^{-2})\right)\\ &= \frac12 + O(x^{-1}). \end{align}$$

And hence

$$\lim_{x\to\infty} \left(\frac{e}{\left(1+\frac1x\right)^x}\right)^x = e^{1/2}.$$