Limit of $n\ln\left(\frac{n-1}{n+1}\right)$ I'm having trouble with this limit:
$$\lim_{n\to ∞}n\ln\left(\frac{n-1}{n+1}\right)$$
It's supposed to be solvable without using l'Hospital's rule. I'm guessing it's a case for the squeeze theorem, but I'm not exactly sure. Any advice?

Quite simply one needs to use the standard limit formula $$\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1\tag{1}$$ We have
\begin{align}
\lim_{n \to \infty}n\log\left(\frac{n - 1}{n + 1}\right) &= \lim_{n \to \infty}n\log\left(1 - \frac{2}{n + 1}\right)\notag\\
&= \lim_{n \to \infty}n\cdot\frac{-2}{n + 1}\cdot\dfrac{\log\left(1 - \dfrac{2}{n + 1}\right)}{-\dfrac{2}{n + 1}}\notag\\
&= -2\lim_{n \to \infty}\frac{n}{n + 1}\cdot\lim_{x \to 0}\frac{\log(1 + x)}{x}\text{ (putting }x = -\frac{2}{n + 1})\notag\\
&= -2\cdot 1\cdot 1 = -2\notag
\end{align}
