What is the fastest way to compute the limit $\lim_{x\to 0^+}x\cosh^{-1}(x\sinh(\frac 1x))$ I would like to compute $$\lim_{x\to 0^+}x\cosh^{-1}(x\sinh(\frac 1x))$$
I used the formula
$$\cosh^{-1}(X)=\ln(X+\sqrt{X^2-1})$$
for any $ X\ge 1$ but i couldn't get a simple equivalent.
Any help or idea will appreciated.
Thanks in advance.
 A: Using the logarithm definition of arccosh yields $$\lim_{x \to 0^+} x\ln\left(x\sinh\left(\frac{1}{x}\right)+\sqrt{x^{2}\sinh^{2}\left(\frac{1}{x}\right)-1}\right)$$
Or breaking up the $\ln$, this is equivalent to $$\lim_{x \to 0^+} \left(x\ln\left(x\right)+x\ln\left(\sinh\left(\frac{1}{x}\right)\right)+x\ln\left(1+\sqrt{1-\frac{1}{x^{2}\sinh^{2}\left(\frac{1}{x}\right)}}\right) \right)$$
You can find the limit of the first part by applying L'Hopital's with $\frac{\ln(x)}{\frac{1}{x}}$ to get $0$. For the third part, make the transformation $x \to \frac{1}{x}$ to get the limit expression is equal to $$\lim_{x \to \infty} \frac{\ln\left(1 + \sqrt{1-\left(\frac{x}{\sinh(x)}\right)^2} \right)}{x}$$
This is $0$ since $\frac{x}{\sinh(x)} \to 0$, so the numerator would go to $\ln(2)$, while the denominator goes to $\infty$.
In a similar way, you can also find the limit of the second part equal to $$\lim_{x \to \infty} \frac{\ln \left( \sinh(x) \right)}{x} = \lim_{x \to \infty} \coth(x) = 1$$
where I applied L'Hopital's rule. Therefore, the limit would be $$0 + 1 + 0 = 1$$
