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?
 A: BIG HINT:
$$\lim_{x\to\infty}\left(\frac{1}{\sinh^{-1}(x)}\right)^{\frac{1}{x}}=$$
$$\lim_{x\to\infty}\exp\left(\ln\left(\left(\frac{1}{\sinh^{-1}(x)}\right)^{\frac{1}{x}}\right)\right)=$$
$$\lim_{x\to\infty}\exp\left(\frac{1}{x}\ln\left(\frac{1}{\sinh^{-1}(x)}\right)\right)=$$
$$\lim_{x\to\infty}\exp\left(\frac{\ln\left(\frac{1}{\sinh^{-1}(x)}\right)}{x}\right)=$$
$$\exp\left(\lim_{x\to\infty}\frac{\ln\left(\frac{1}{\sinh^{-1}(x)}\right)}{x}\right)=$$
$$\exp\left(-\lim_{x\to\infty}\frac{\ln\left(\sinh^{-1}(x)\right)}{x}\right)=$$
$$\exp\left(-\lim_{x\to\infty}\frac{\frac{\text{d}}{\text{d}x}\left(\ln\left(\sinh^{-1}(x)\right)\right)}{\frac{\text{d}}{\text{d}x}\left(x\right)}\right)=$$
$$\exp\left(-\lim_{x\to\infty}\frac{\frac{1}{\sinh^{-1}(x)\sqrt{1+x^2}}}{1}\right)=$$
$$\exp\left(-\lim_{x\to\infty}\frac{1}{\sinh^{-1}(x)\sqrt{1+x^2}}\right)=$$
$$\exp\left(-\frac{1}{\lim_{x\to\infty}\sinh^{-1}(x)\sqrt{1+x^2}}\right)=$$
$$\exp\left(-\frac{1}{\left(\lim_{x\to\infty}\sinh^{-1}(x)\right)\left(\lim_{x\to\infty}\sqrt{1+x^2}\right)}\right)=$$
$$\exp\left(-\frac{1}{\left(\lim_{x\to\infty}\sinh^{-1}(x)\right)\left(\sqrt{\lim_{x\to\infty}x^2}\right)}\right)$$
A: 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
$$
y=\log\bigl(x+\sqrt{1+x^2}\bigr)\sim\log(2\,x)\text{ as }x\to\infty.
$$
