The limit of a function How can I prove that the limit of $\left(\frac{1}{\operatorname{arcsinh}(x)}\right)^\frac{1}{x}$ is $1$ as $x$ goes to $\infty$?
So how do I see that the arcsinh grows so slowly that the exponential reduces it to one? Is it (arcsinh) growing logarithmically or polynomially?

Let $y=\sinh^{-1} x$. Then
$$
\frac{e^y-e^{-y}}{2}=x\implies e^{2y}-2\,x\,e^y-1=0.
$$
Solving for $y$ we get
$$
y=\log\bigl(x+\sqrt{1+x^2}\bigr)\sim\log(2\,x)\text{ as }x\to\infty.
$$
