Question: I just can't get past the $x-1$. The $2$ I know how to get rid of.

$$\lim\_{x\to +∞}\frac {\ln\left((x-1)^2\right)}{x} $$

Answer: 
Use *equivalents*:

$(x-1)^2\sim\_\infty x^2$ and none of them approach $1$ in a neighbourhood of $+\infty$, hence $\;\ln(x-1)^2\sim\_\infty \ln x^2=2\ln x\;$ and finally
$$\frac{\ln(x-1)^2}{x}\sim\_\infty2\,\frac{\ln x}x \to 0.$$

*Note:* In general equivalence of functions is not compatible with composition on the left with another function.
By applying [L'Hôpital's rule](https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule) you have: 
$$\lim\_{x\rightarrow \infty} (\ln(x-1)^2)'=\lim\_{x\rightarrow \infty} \frac{2}{x-1}=0$$
and $$\lim\_{x\rightarrow \infty}(x)'=1$$
Therefore the limit $$\lim\_{x\rightarrow \infty}\frac{(\ln(x-1)^2)'}{x'}=\frac{0}{1}=0$$
exists and it is equal to
$$\lim\_{x\to +∞}\frac {ln(x-1)^2}{x}$$