Calculate limit without L'Hopital I need a some help with this.

Calculate:

$$\lim_{x\to\infty}\frac{\ln(x-1)}x$$

I know the answer is zero. But dont know how to handle the $\ln(x-1)$
You have that $$\log \sqrt{x -1} &lt; \sqrt{x - 1} \Rightarrow \log (x-1) &lt; 2\sqrt{x-1} \Rightarrow \frac{\log (x-1)}{x} &lt; 2\frac{\sqrt{x-1}}{x}$$

Can you take it from here?
Well, if it's infinity over infinity you can still use L'Hopital's rule.

the derivative of both is:

( 1/(x - 1 ) ) / 1

invert and multiply and then plugin infinity in for X.

The denominator is getting infinitely large and therefore we can assume it's getting closer and closer to 0. Which is why it's 0 (or at least that's how my instructor taught me)