L'Hôpital's as $x$ tends to infinity I'm searching for the explanation to the limit of:
$$
\lim\limits_{x\to\infty} x\, \ln\frac{x+1}{x-1}.
$$
I know the answer is 2, but I can't seem to get there.
The problem is in my textbook under a section with l'Hôpital.
 A: Hint: $$\ln \frac{x+1}{x-1} = \ln(x+1) - \ln(x-1)$$ so that $$x \cdot \ln \frac{x+1}{x-1}$$ can be rewritten as $$\frac{\ln(x+1) - \ln(x-1)}{x^{-1}}.$$ Now,
$$ \lim_{x \to \infty} x \ln \frac{x+1}{x-1} = \lim_{x \to \infty} \frac{\frac{1}{x+1}-\frac{1}{x-1}}{\frac{-1}{x^2}}$$
Find a common denominator for the two terms in the numerator, cancel fractions, and you will arrive at the desired answer of 2.
Note: Indeed, $\ln \frac{x+1}{x-1} \rightarrow 0$ and $ x^{-1} \rightarrow 0$ as $x \rightarrow \infty$. Therefore, we have the required form $0/0$ and L'Hopital's rule is applicable.
A: You don't need L'Hospital to solve it. You can write your limit as 
$$\lim\limits_{x\to\infty} x\ln\left(\frac{x+1}{x-1}\right)=\lim\limits_{x\to\infty} \ln\left(\frac{x+1}{x-1}\right)^x=\ln \lim\limits_{x\to\infty} \left(\frac{x+1}{x-1}\right)^x.$$
Now, 
$$\lim\limits_{x\to\infty} \left(\frac{x+1}{x-1}\right)^x$$
is an indetermination $1^{\infty}.$ This can be solved as 
$$\lim\limits_{x\to\infty} \left(\frac{x+1}{x-1}\right)^x=e^{\lim\limits_{x\to\infty} \left(\frac{x+1}{x-1}-1\right)x}.$$
Anyway, if you want to apply L'Hospital you can express your limit as
$$\lim\limits_{x\to\infty} x\ln\left(\frac{x+1}{x-1}\right)=\lim\limits_{x\to\infty} \frac{\ln\frac{x+1}{x-1}}{\frac{1}{x}}$$ to get an indetermination $\frac{0}{0}.$ Now you can apply L'Hospital. That is,
$$\lim\limits_{x\to\infty} \frac{\ln\frac{x+1}{x-1}}{\frac{1}{x}}=\lim\limits_{x\to\infty} \frac{\frac{x-1}{x+1} \frac{-2}{(x-1)^2}}{\frac{-1}{x^2}}=\cdots =2.$$ 
A: Way 1.
$$
x \ln\frac{x+1}{x-1}=x\ln\left(1+\frac{2}{x-1}\right)=x\left(\frac{2}{x-1}+\mathcal O\left(\frac{1}{x^2}\right)\right)=\frac{2}{x-1}+O\left(\frac{1}{x}\right)\to 2.
$$
We have used the fact that
$$
\ln (1+w)=w+{\mathcal O}(w^2),
$$
for $w$ small.
Way 2. Otherwise, set $h=1/x$, and now $h\to 0$, and our linear becomes
$$
\lim_{h\to 0}\frac{\ln\left(\frac{1+h}{1-h}\right)}{h},
$$
but this is simply equal to $f'(0)$ where
$$
f(x)=\ln\left(\frac{1+x}{1-x}\right)=\ln(1+x)-\ln(1-x)
$$
and clearly
$$
f'(x)=\frac{1}{x+1}+\frac{1}{1-x},
$$
and thus $f'(0)=2$.
A: You can observe that
$$
\frac{x+1}{x-1}=\frac{x-1+2}{x-1}=1+\frac{2}{x-1}
$$
Now write $x=(x-1)+1$ and transform your limit in
$$
\lim_{x\to\infty}\left((x-1)\log\left(1+\frac{2}{x-1}\right)+
  \log\left(1+\frac{2}{x-1}\right)\right)
$$
The second summand has limit $0$, so it can be removed and you can do the substitution $1/t=x-1$ that brings the limit into
$$
\lim_{t\to0^+}\frac{\log(1+2t)}{t}
$$
that's the derivative of $f(t)=\log(1+2t)$ at $0$; since
$$
f'(t)=\frac{2}{1+2t}
$$
we have $f'(0)=2$ as in your solution sheet.
For another method, more direct, observe that
$$
\lim_{x\to\infty}\log\frac{x+1}{x-1}=0
$$
(and that $(x+1)/(x-1)>0$ for $x>1$), so your limit is in the form $\infty\cdot0$. Such a form becomes either $0/0$ or $\infty/\infty$ by “reversing” one of the factors. More precisely, if $\lim_{x\to?}f(x)=\infty$ and $\lim_{x\to?}g(x)=0$, you can consider either
$$
f(x)g(x)=\frac{g(x)}{1/f(x)}
$$
which takes the $0/0$ form, or
$$
f(x)g(x)=\frac{f(x)}{1/g(x)}
$$
which is in the form $\infty/\infty$. We can apply l'Hôpital's theorem to both forms. I used ? to mean any point (finite or infinite) at which the limit is taken.
Doing $1/\log(\dots)$ is a last resource, because the derivative won't lose the logarithm; on the other hand, the derivative of the logarithm loses it, while the derivative of $1/x$ is “easy”, so the first strategy to try is with the $0/0$ form above:
$$
\lim_{x\to\infty}x\log\frac{x+1}{x-1}=
\lim_{x\to\infty}\frac{\log\frac{x+1}{x-1}}{1/x}
$$
Now the derivative of the numerator $\log\frac{x+1}{x-1}=\log(x+1)-\log(x-1)$ is
$$
\frac{1}{x+1}-\frac{1}{x-1}=\frac{(x-1)-(x+1)}{x^2-1}=-\frac{2}{x^2-1}
$$
and the derivative of the denominator $1/x$ is $-1/x^2$, so we have
$$
\lim_{x\to\infty}\frac{\log\frac{x+1}{x-1}}{1/x}
\overset{\mathrm{(H)}}{=}
\lim_{x\to\infty}\frac{-2/(x^2-1)}{-1/x^2}\overset{*}{=}
\lim_{x\to\infty}\frac{2x^2}{x^2-1}=2
$$
An equality is marked with (H) to show the application of l'Hôpital's theorem; the one marked with $*$ shows that it's very important to simplify the formula before attempting again with the big theorem.
A: Using the expression in mfl's answer (using continuity of the logarithmic function):
$$\lim_{x\to\infty}x\log\frac{x+1}{x-1}=\log\lim_{x\to\infty}\left(1+\frac2{x-1}\right)^x$$
But
$$\lim_{x\to\infty}\left(1+\frac2{x-1}\right)^x=\lim_{x\to\infty}\left(1+\frac2{x-1}\right)^{x-1}\left(1+\frac2{x-1}\right)= e^2\cdot1=e^2$$
and thus the wanted limit is
$$\log e^2=2$$
