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: 
$$\lim\_{x\to \infty} \frac{\ln(x-1)^2}{x}$$

It is not hard to show $\ln x< x$

$\ln(x-1)^2=4\ln\sqrt{x-1}\leq 4\sqrt{x-1}\leq 4\sqrt x$

So the limit is smaller than

$$\lim\_{x\to \infty} \frac{4\sqrt{x}}{x}=\lim\_{x\to \infty} \frac 4{\sqrt x}=0$$

Now the following is true because square root must be non negative

$-4\sqrt x\leq ln(x-1)^2$

So the limit is bounded by $0$
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
$$\lim\_{x\to +∞}\frac {ln(x-1)^2}{x}$$