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?
 A: Hint:
Set $\dfrac1x=h$  to find
$$\lim_{h\to0^+}\dfrac{\ln\dfrac{1+h}{1-h}}h =\lim_{h\to0^+}\dfrac{\ln(1+h)}h+\lim_{h\to0^+}\dfrac{\ln(1-h)}{-h}=?$$
A: 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 $\infty$.
A: Your first reasoning is wrong, because you have $0\cdot\infty$ uncertainty. Second way is correct with correct answer.
One more possible way:
$$x\ln\left(\frac{x+1}{x-1}\right)=x\ln\left(\frac{x-1+2}{x-1} \right)=x\ln\left(1+\frac{2}{x-1} \right)\sim \frac{2x}{x-1}\to 2$$
A: $\lim\limits_{x\to \infty} x = 0$ and $\lim\limits_{x\to \infty}\ln {\frac {x+1}{x-1}} = 0$ so you have "$0\cdot \infty$" type indeterminate.
Try L'Hopital
$\lim x{\ln\frac {x+1}{x-1}}=$
$\lim \frac{\ln\frac{x+1}{x-1}}{\frac 1x}=$
$\lim \frac{\frac {x-1}{x+1}\cdot \frac {(x-1)-(x+1)}{(x-1)^2}}{-\frac 1{x^2}}=$
$\lim -x^2 \frac {-2}{(x+1)(x-1)}=$
$\lim \frac {2x^2}{x^2 -1} = 2$
A: To prove ur reasoning is wrong, consider a simpler problem: $\lim\limits_{x\to \infty} x \cdot \frac{1}{x}$. here if I say $\frac{1}{x}$ goes to zero and $x$ goes to infinity then the limit is infinity how would that reasoning make sense? cancel the $x$'s and u can see the limit is $1$.
Coming to you problem.
Why do you assume if $\ln {\frac {x+1}{x-1}}$ tends to 0 and you multiply by x you get infinity? $0*\infty$ is indeterminate like the above limit...  and you must convert this into a form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ so we can apply l'hopital.
Solution:
$\lim\limits_{x\to \infty} x \ln {\dfrac {x+1}{x-1}} =\lim\limits_{x\to \infty} \dfrac{(\ln \dfrac {x+1}{x-1})'}{(\dfrac{1}{x})'} =$
$\lim\limits_{x\to \infty} -x^2\cdot \dfrac 1{\dfrac {x+1}{x-1}}(\frac {x+1}{x-1})' =$
$\lim\limits_{x\to \infty}  -x^2\cdot \dfrac {x-1}{x+1}\dfrac {(x+1)'(x-1) - (x+1)(x-1)'}{(x-1)^2}=$
$\lim\limits_{x\to\infty}-x^2\cdot \dfrac 1{x+1}\dfrac {(x-1)-(x+1)}{x-1}=$
$\lim\limits_{x\to \infty} \dfrac {2x^2}{x^2-1} = \lim\limits_{x\to \infty} \dfrac {4x}{2x} =2$.
A: $f(x):=\log \left (\dfrac{ (x+1)^x}{(x-1)^x} \right)=$
$\log \left (\dfrac {(1+1/x)^x}{(1-1/x)^x} \right)=$
$\lim_{x \rightarrow \infty} f(x)= \log \dfrac {e^1}{e^{-1}}=2.$
Used:
$1)\log$ is a continuos function;
$2)\lim_{x \rightarrow \infty} (1+a/x)^x= e^a$, $a$ real.
A: My attempt
$\lim\limits_{x\to \infty} xln(\frac{x+1}{x-1})=\lim\limits_{x\to \infty} xln(x+1)- xln(x-1) $
Suppose $t=x+1$
So $x=t-1$
$\lim_{t\to \infty}tln(t) - ln(t) -tln(t-2)+ln(t-2)$
$=lim_{t\to \infty} tln(\frac{t}{t-2})+ln(\frac{t-2}{t})$
$\lim\limits_{t\to \infty} (t(\frac{t} {t-2}-1)\frac{ln(\frac{t} {t-2})}{(\frac{t} {t-2}-1)}+ln(\frac{t-2}{t})=\lim\limits_{t\to \infty} ((\frac{2t} {t-2})\frac{ln(\frac{t} {t-2})}{(\frac{t} {t-2}-1)}+ln(\frac{t-2}{t})$
Suppose $ w=\frac{t} {t-2}$
$\lim_{w\to1}2w(\frac{ln(w)}{w-1})+ln(\frac{1}{w})=2$
