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?
 A: You can write
$$
\frac{n-1}{n+1}=\frac{n+1-2}{n+1}=1-\frac{2}{n+1}
$$
so you can also write your limit as
$$
\lim_{n\to\infty}\frac{n}{n+1}(n+1)\ln\left(1-\frac{2}{n+1}\right)
=
\lim_{n\to\infty}\frac{n}{n+1}
\ln\biggl(\biggl(1-\frac{2}{n+1}\biggr)^{n+1}\biggr)
$$
Now you should know that
$$
\lim_{n\to\infty}\biggl(1+\frac{a}{n}\biggr)^n=e^a
$$
A: \begin{align}
&\lim_{n\to \infty}n\ln\left(\frac{n-1}{n+1}\right)=\lim_{n\to \infty}\ln\left(\frac{n+1-2}{n+1}\right)^n\\
&=\lim_{n\to \infty}\ln\left(1-\frac{2}{n+1}\right)^n\\
&=\lim_{n\to \infty}\ln\left(1-\frac{2}{n+1}\right)^{\frac{-(n+1)}{2}\cdot\frac{-2n}{n+1}}\\
&=\lim_{n\to \infty}\ln\left(\left(1-\frac{2}{n+1}\right)^{\frac{-(n+1)}{2}}\right)^{\frac{-2n}{n+1}}\\
&=\lim_{n\to \infty}\ln \left(e^{\frac{-2n}{n+1}}\right)=\lim_{n\to \infty}\frac{-2n}{n+1}=-2
\end{align}
A: With equivalents: $\;\ln(1-u)\sim_0 -u$, hence
$$n\ln\Bigl(\frac{n-1}{n+1}\Bigr)=n\ln\Bigl(1-\frac{2}{n+1}\Bigr)\sim_\infty-\frac{2n}{n+1}\sim_\infty -2.$$
A: In THIS ANSWER, I showed using standard tools that don't rely on derivatives that, for $z>-1$, the logarithm function satisfies the inequalities 
$$\frac{z}{z+1}\le \log (1+z)\le z$$
Now, we note that $\log \left(\frac{n-1}{n+1}\right)=\log \left(1-\frac{2}{n+1}\right)$.  Therefore, we have for $n>1$
$$-\frac{2n}{(n-1)}\le n\log \left(1-\frac{2}{n+1}\right)\le -\frac{2n}{n+1}$$
Using the Squeeze Theorem, we obtain the limit
$$\lim_{n\to \infty}n\,\log \left(\frac{n-1}{n+1}\right)=-2$$
A: 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, 
$$=\lim_{t\to 0}\frac{-\left(t+\frac{t^2}{2}+O(t^2)\right)-\left(t-\frac{t^2}{2}+O(t^2)\right)}{t}$$
$$=\lim_{t\to 0}\left(-\left(1+\frac{t}{2}+O(t)\right)-\left(1-\frac{t}{2}+O(t)\right)\right)$$
$$=-\left(1+0\right)-\left(1-0\right)=\color{red}{-2}$$
A: 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}
