Limit of arccosh x - ln x as x -> infinity

[GwtBc]

Homework Statement

find the limit of arccoshx - ln x as x -> infinity

Homework Equations

$arccosh x = \ln (x +\sqrt[]{x^2-1} )$

The Attempt at a Solution

$\lim_{x \to \infty }(\ln (x + \sqrt{x^2-1} ) - \ln (x)) = \lim_{x \to \infty} \ln (\frac{x+\sqrt{x^2-1}}{x}) \ln (1 + \lim_{x \to \infty}\frac{\sqrt{x^2-1}}{x})$

I can see that the limit of the second part is also going to one, but I can't manipulate the expression to show this.

 

[Ssnow]

I think it is $\lim_{x\rightarrow +\infty} \ln{(x+\sqrt{x^{2}-1})}-\ln{x}=\lim_{x\rightarrow +\infty} \ln{\frac{x+\sqrt{x^{2}-1}}{x}}$ that is $\lim_{x\rightarrow +\infty} \ln{\left(1+\sqrt{1-\frac{1}{x^{2}}}\right)}$

 

[GwtBc]

I was just about to come here and say "Wow I'm really stupid, I got it now". But you'd already answered. Thanks :D.