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I have to show that

$$\arccos(x) = \sqrt{2-2x} + O((1-x)^{3/2}) \\ \text{arcosh}(x) = \sqrt{2x-2} + O((x-1)^{3/2})$$

for $x\to 1^-$ ($x$ approaching 1 from below), $\text{arcosh}(x)$ denotes the inverse of $\cosh(x). $ I tried rewriting it as $\arccos(x) - \sqrt{2-2x} = O((1-x)^{3/2}) $ and then plugging it into the definition, but that didn't quite help since everything got too messy. I don't know what I could rewrite $\arccos(x)$ as either. Any hints?

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    Write $x = 1- \varepsilon$ [$1 + \varepsilon$ for the hyperbolic case] and ask yourself for which $\delta$ you have $\cos \delta = 1 - \varepsilon$. – 2017-01-23
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    you can also investigate the asymptotics of $\sqrt{1\pm x^2}^{-1}$ – 2017-01-23

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Hint. One may observe that, for $x>1$, one has $$ \begin{align} \text{arcosh}(x)&=\log\left(x+\sqrt{x^2-1} \right) \end{align} $$ giving, as $x \to 1^+$, by the use of a Taylor series expansion, as $\varepsilon \to 0^+$, with $x=1+\varepsilon$, $$ \begin{align} \text{arcosh}(x)&=\log\left(1+\varepsilon+\sqrt{(1+\varepsilon)^2-1} \right) \\&=\log\left(1+\varepsilon+\sqrt{2\varepsilon+\varepsilon^2} \right) \\&=\left(\varepsilon+\sqrt{2\varepsilon+\varepsilon^2} \right)-\frac12\left(\varepsilon+\sqrt{2\varepsilon+\varepsilon^2} \right)^2+O(\varepsilon^{3/2}) \\&=\sqrt{2} \sqrt{\varepsilon}+O(\varepsilon^{3/2}) \\&=\sqrt{2} \sqrt{x-1} + O((x-1)^{3/2}) \end{align} $$ as announced.

The same approach applies to $\text{arcos}$.

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    Awesome, thanks! – 2017-01-23
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    @lappen68 You are welcome. – 2017-01-23
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    Does one have a similar fomula for arccos? – 2017-01-23
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    @lappen68 Yes, one may use $\arccos(x)=- \mathrm{i} \log \left( x+\sqrt{x^2-1} \right)$. Please, have a look here: https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Logarithmic_forms – 2017-01-23
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    Alright, I actually thought about that one, however that $i$ confused me a bit. Thanks again! – 2017-01-23