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?

You have done it correctly but what you said in the beginning is not correct. If you mulitiply two functions one of which has limit $0$ and the other has limit $\infty$ you can say nothing about the limit of the product.
Examples: As $x \to \infty$, $\frac  1 x \to 0$ and $\sqrt x \to \infty$ but the product tends to $0$; On the other hand $\frac  1 x \to 0$ and $x^{2} \to \infty$ but the product tends to $\infty$.
