Find the following limit,
$$\lim_{x\to {\infty}} x \ln\left(\frac{x+1}{x-1} \right)$$
I tried this way, that is $$\lim_{x\to {\infty}} x\ln\left(\frac{1+\frac{1}{x}}{1-\frac{1}{x}} \right)=\infty \times \ln1= \infty \times 0=0$$
Noticed my logic is horribly wrong. Tested it out on a calculator and the limit should be near $0.8$.
My second try involves $$\lim_{x\to {\infty}} x\ln\left(\frac{x+1}{x-1} \right)=\lim_{x\to {\infty}} x(\ln(x+1)-\ln(x-1))$$
Where I still have no idea how to proceed.
Any hints? Thanks in advance! List them as solutions. I am looking for hints only.