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?

Notice, let $\frac{1}{n}=t$, hence $$\lim_{n\to \infty}n\ln\left(\frac{n-1}{n+1}\right)$$ $$=\lim_{t\to 0}\frac{1}{t}\ln\left(\frac{\frac{1}{t}-1}{\frac{1}{t}+1}\right)$$ $$=\lim_{t\to 0}\frac{\ln\left(\frac{1-t}{1+t}\right)}{t}$$
$$=\lim_{t\to 0}\frac{\ln\left(1-t\right)-\ln\left(1+t\right)}{t}$$
Now, using Taylor's series, 
$$=\lim_{t\to 0}\frac{-\left(t+\frac{t^2}{2}+O(t^2)\right)-\left(t-\frac{t^2}{2}+O(t^2)\right)}{t}$$
$$=\lim_{t\to 0}\left(-\left(1+\frac{t}{2}+O(t)\right)-\left(1-\frac{t}{2}+O(t)\right)\right)$$
$$=-\left(1+0\right)-\left(1-0\right)=\color{red}{-2}$$
