Find the following limit, $\lim_{x\to {\infty}} x\ln((x+1)/(x-1))$ 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. 

Update: Upps, noticed too late, that only hints were wished. I can't make the solution after the first reformulation at the lhs (which were the hint then) invisible, don't know the latex-trick - sorry; somebode else may edit?
If I'm not too dense at the moment I think it is
 $$ \begin{eqnarray} \lim_{x \to \infty } \ln\left( \left( 1+{2\over x-1 }\right)^x\right)
 &=&\lim_{x \to \infty } &\ln\left(  \left( 1+{2\over x }\right)^{x+1}  \right) \\
 &=&\lim_{x \to \infty } &\ln \left(  \left( 1+{2\over x }\right)^x \left( 1+{2\over x }\right) \right) \\
 &=& &\ln \left(  e^2 \left( 1+ 0\right) \right) \\
 &=& 2 
 \end{eqnarray}$$
