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 $\ln(x-1)$

We will prove instead that $\frac{\log x}{x} \to 0$, and leave you to deduce what you need. Suppose $x \geq e$ (here $e$ is the base of the logarithm). Then
$$\frac{\log (ex)}{ex} = \frac{\log x + 1}{ex} \leq \frac{2}{e} \frac{\log x}{x}, $$
since $e \leq x$ implies $1 \leq \log x$ (since $\log x$ is monotone). Monotonicity of $\log x$ and $x$ shows that for $e \leq x \leq e^2$ we have
$$ \frac{\log x}{x} \leq \frac{\log (e^2)}{e} = \frac{2}{e}. $$
Take now an arbitrary $x$ and divide it by $e$ enough times so that you obtain something in the range $[e,e^2]$. You can express the result by $x = e^n t$, where $e \leq t \leq e^2$ and $n \geq 0$. The above two calculations show that
$$ \frac{\log x}{x} \leq \left(\frac{2}{e}\right)^{n+1}. $$
As $x\to\infty$, the corresponding power $n$ also tends to infinity, and so the ratio $\frac{\log x}{x}$ tends to zero.
