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Does $\tanh(x)$ have an asymptotic expansion for $x \rightarrow \infty$?

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    Then my first comment applies. Since $\exp\frac1{x}$ has an essential singularity at $x=0$, an asymptotic expansion of the form $c_0+\frac{c_1}{x}+\frac{c_2}{x^2}+\dots$ can't be done.2011-08-16

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Write

$\tanh(x) = \frac{1-e^{-2x}}{1+e^{-2x}}$

Then you get an asymptotic expansion with powers of $e^{-2x}$ (which goes to $0$ as $x$ goes to infinity). It starts as

$\tanh(x) = 1 - 2 e^{-2x} + o(e^{-2x})$

There is no asymptotic expansion with powers of $x$ as that would imply that $e^{-x}$ has one (remember we're talking about asymptotic expansion when $x$ goes to infinity).

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    @Keith: Glad to help.2011-08-20
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Using the definition of $\tanh(x)$, $ \begin{align} \tanh(x)&=\frac{e^x-e^{-x}}{e^x+e^{-x}}\\ &=\frac{1-e^{-2x}}{1+e^{-2x}}\\ &=\frac{2}{1+e^{-2x}}-1\\ &=1-2e^{-2x}+2e^{-4x}-2e^{-6x}+2e^{-8x}-\dots \end{align} $ This converges to $\tanh(x)$ for all $x>0$. It also describes how $\tanh(x)$ behaves as $x\to\infty$.