<s>[INST] What is? $ \lim_{x \to \infty} \left(\frac{e}{\left( 1 + \frac {1}{x} \right)^x} \right)^x $ 
Find $$ \lim_{x \to \infty} \left( \dfrac{e}{\left( \left( 1 + \frac {1}{x} \right)^x \right)} \right)^x. $$


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