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 $\ln(x-1)$

Because $ \ln{x-1} $ is a monotonically increasing function in its domain $]1,\infty]$:
$$ 0 < \ln{x-1} < \ln{x} \Leftrightarrow 0 < \frac{\ln{x-1}}{x} < \frac{\ln{x}}{x}$$ 
Our strategy from here is to prove that the right-hand side ($ln(x)/x$) is approaching $0$ as $x$ is approaching infinity. We will then use the Squeeze theorem to prove that the target expression is approaching 0 as well.
Let's proceed by introducing the substitution $x=e^{t}$ and then utilize a standard limit in the last step:
$$\lim_{x\to\infty} \frac{\ln{x}}{x} = \lim_{t\to\infty} \frac{\ln{e^t}}{e^{t}} = \lim_{t\to\infty} \frac{t}{e^t} = 0.$$
Using the squeeze theorem, we can now conclude that:
$$\lim_{x\to\infty} \frac{\ln{x-1}}{x} = 0.$$
