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Can someone show me how to calculate the limit:

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

I tried to use taylor series but failed.

Thanks

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2 Answers 2

up vote 6 down vote accepted

Take the logarithm,

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

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Assume the limit exists and has value $L$ Then

$$\log{L} = \lim_{x\to\infty} x \log{\left [ \frac{1}{e} \left ( 1+\frac{1}{x}\right)^x\right]} =\lim_{x\to\infty} x \left [ -1 + x \log{\left ( 1+\frac{1}{x}\right)}\right]$$

Taylor expand the log term for large $x$:

$$\log{L} = \lim_{x\to\infty} x \left [-1 + x \left (\frac{1}{x} - \frac{1}{2 x^2} + \cdots\right)\right]= -\frac12$$

Therefore $L=e^{-1/2}$.

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