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. 
 A: 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
 $$ \begin{eqnarray} \lim_{x \to \infty } \ln\left( \left( 1+{2\over x-1 }\right)^x\right)
 &=&\lim_{x \to \infty } &\ln\left(  \left( 1+{2\over x }\right)^{x+1}  \right) \\
 &=&\lim_{x \to \infty } &\ln \left(  \left( 1+{2\over x }\right)^x \left( 1+{2\over x }\right) \right) \\
 &=& &\ln \left(  e^2 \left( 1+ 0\right) \right) \\
 &=& 2 
 \end{eqnarray}$$
A: Hint: Our expression is equal to
$$\ln\left(\left(\frac{x+1}{x-1}\right)^x\right).$$
Note that 
$$\frac{x+1}{x-1}=1+\frac{2}{x-1}.$$
Another approach that will work is to rewrite the expression as
$$\frac{\ln\left(\frac{x+1}{x-1}\right)}{\frac{1}{x}},$$
use L'Hospital's Rule once, and then a little algebra. 
Remark: Any solution that attempts to calculate "$\infty \cdot 0$" is automatically incorrect. 
A: Rewrite it as $$x \ln \left(\dfrac{1+1/x}{1-1/x}\right) = x \left( \ln(1+1/x) - \ln(1-1/x)\right) = x \left( \left( \dfrac1x - \dfrac1{2x^2} + \dfrac1{3x^3} - \cdots\right) - \left( -\dfrac1x - \dfrac1{2x^2} - \dfrac1{3x^3} - \cdots\right)\right) = 2 + \dfrac2{3x^2} + \cdots$$
A: $0 \cdot \infty$ is not $0$ is an undeterminate form. Write it as fraction, move one of them to the denominator (careful it matters which one).
Another small hint for later: when you need to derivare the $\ln$ of a  fraction, what is the $\ln (\frac{a}{b})$?
