How do I solve $\lim_{x\to +\infty}\frac {\ln(x-1)^2}{x} $?

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 1

$$\lim_{x\to \infty} \frac{\ln(x-1)^2}{x}$$

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 $0$

Answer 2

By applying L'Hôpital's 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 $$\lim_{x\to +∞}\frac {ln(x-1)^2}{x}$$

Answer 3

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.