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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?

amWhy
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jujuju
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1 Answers1

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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

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

Daniel Fischer
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  • log=>ln? why ln(1+x^(-1))=1/x-1/(2x^2)+o(x^-3) – jujuju Aug 20 '13 at 23:16
  • I still dont know Why is this incorrect? – jujuju Aug 20 '13 at 23:39
  • @jujuju $\log$ is indeed the natural logarithm (mathematicians rarely use $\ln$). The Taylor expansion of $\log (1+y)$ is $$\log (1+y) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}y^n,$$ cutting off after the quadratic term gives the used form (with $y = 1/x$). Your result was incorrect because you can't separate the limit into two steps here where the exponent doesn't have a finite limit (and of course $1^x = 1$ for all $x$). When you have $\lim\limits_{x\to\infty} f(x)^{g(x)}$, you can separate the limits if both $f(x)$ and $g(x)$ have a finite limit, but not if $g(x) = x$. – Daniel Fischer Aug 21 '13 at 00:00