Evaluating a limit regarding arccosh

Question

The problem reads out as:

Evaluate the following limit

$$ \lim_{x\to\infty} (\textrm{arccosh} x)^\frac{1}{x} $$

Now I'm pretty sure the professor is looking for me to use L Hopitals rule, however, I'm not sure how to get rid of the exponent that is specifically outside of the parentheses. Since I know what arccosh is equal to, I put it back inside of the limit but am not sure what to do with the exponent, since from what I understand of the exponent rule for log, you can only move the exponent to the front of the ln if it's inside the parentheses. I'm aware of the power rule for limits as well, but I'm not quite sure how it would help me. I've brought this to a friend who is a math major but he hasn't taken calculus in a while and he didn't know and the tutor at my university didn't know either (which was pretty concerning to me).

Arccosh: $$\textrm{arccosh} = \ln(x+\sqrt{x^2-1})$$

How far I get: $$ \lim_{x\to\infty} (\ln(x+\sqrt{x^2-1}))^\frac{1}{x} $$

Answer 1

Use that $$\begin{align} \lim_{x\to a} f(x)^{g(x)} &= \lim_{x\to a} \exp \bigg( \log \big( f(x)^{g(x)} \big) \bigg) \\ &= \lim_{x\to a} \exp \bigg( g(x) \log f(x) \bigg) \\ &= \exp \bigg( \lim_{x\to a} g(x) \log f(x) \bigg) \end{align}$$ and rewrite the latter so that you can now apply L'Hôpital's rule.

Answer 2

For large $x$, $\text{arcosh}x\sim\ln x$ and $$\lim_{x\to \infty}(\ln x)^{1/x}=1$$

because $\dfrac{\ln(\ln x))}x\to0$.

Answer 3

hint

Put $x=\cosh y$ and compute

$$\displaystyle{\lim_{y\to+\infty}y^{\frac{1}{\cosh(y)}}=\lim_{y\to+\infty}e^{\frac{2}{e^y+e^{-y}}\ln(y)}}$$

using $$\lim_{y\to+\infty}\frac{\ln(y)}{e^y}=0$$