Finding $\lim_{x\rightarrow \infty}x\ln\left(\frac{x+1}{x-1}\right)$ I'm trying to find out :
$$\lim_{x\rightarrow \infty}x\ln\left(\frac{x+1}{x-1}\right)$$
First I tried by inspection. I suspect that
$$\lim_{x\rightarrow \infty} \left( \frac{x+1}{x-1}\right)=1$$
So I expect $\ln (\cdots)\rightarrow 0$ as $x\rightarrow \infty$. While If I multiply this by $x$, One might expect this to go infinite.

Next what I have done is to use a substitution, I don't know if it's right to do or not.
$$x'=\frac{x+1}{x-1}\Rightarrow x=\frac{x'+1}{x'-1}$$
So that
$$\lim_{x\rightarrow \infty}x\ln\left(\frac{x+1}{x-1}\right)\Rightarrow \lim_{x'\rightarrow 1}\left(\frac{x'+1}{x'-1}\right)\ln x'$$
Now I can use L'Hospital rule
$$\lim_{x'\rightarrow \infty}\frac{\ln x'+\frac{x'+1}{x'}}{1}=2$$

Can I use such a substitution always to solve problems? What's wrong with the reasoning I did earlier?

Hint:
Set $\dfrac1x=h$  to find
$$\lim_{h\to0^+}\dfrac{\ln\dfrac{1+h}{1-h}}h =\lim_{h\to0^+}\dfrac{\ln(1+h)}h+\lim_{h\to0^+}\dfrac{\ln(1-h)}{-h}=?$$
