Compute the limit of $1 - n \ln \left(\dfrac{2n + 1}{2n - 1}\right)$ 
Compute the limit: $$\lim_{n \to \infty} \left(1 - n \ln \left(\dfrac{2n + 1}{2n - 1}\right)\right) $$

Can someone help me to solve this limit? I forgot how to manupulate fractions in limit calculus.
 A: Use the fact that: $\displaystyle\lim_{x\to 0} \frac{\ln(1+x)}{x}=1$.
Hence 
$$\lim_{n\to\infty}\left(1-n\ln\left( \frac{2n + 1}{2n - 1}\right)\right)=1-\lim_{n\to\infty}n\cdot\ln\left(1+ \frac{2}{2n - 1}\right)\\
=1-\lim_{n\to\infty}n\cdot\frac{2}{2n - 1}\cdot\frac{\ln\left(1+ \frac{2}{2n - 1}\right)}{\frac{2}{2n - 1}}=1-1\cdot\lim_{n\to\infty}\frac{2n}{2n - 1}=1-1=0.$$
A: 
PRIMER:
In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$\frac{x-1}{x}\le \log(x)\le x-1 \tag 1$$

We present an approach the relies on the inequalities in $(1)$ along with the squeeze theorem.  Note from $(1)$ that
$$\frac{2}{2n+1}\le \log\left(\frac{2n+1}{2n-1}\right)\le \frac{2}{2n-1} \tag 2$$
Then, using $(2)$ reveals
$$-\frac{1}{2n-1}\le 1-n\log\left(\frac{2n+1}{2n-1}\right)\le \frac{1}{2n+1}$$
whereupon applying the squeeze theorem, we obtain the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\left(1-n\log\left(\frac{2n+1}{2n-1}\right)\right)=0}$$

A: You can write it as:
$L = \displaystyle \lim_{n \to \infty} \dfrac{\dfrac{1}{2n} - \dfrac{1}{2}\ln\left(\dfrac{1+\dfrac{1}{2n}}{1-\dfrac{1}{2n}}\right)}{\dfrac{1}{2n}}= \displaystyle \lim_{x \to 0} \dfrac{x-\dfrac{\ln(1+x) - \ln(1-x)}{2}}{x}= 1- \dfrac{\displaystyle \lim_{x \to 0}\dfrac{\ln(1+x)-\ln(1-x)}{x}}{2}= 0$ by L'hospitale rule.
A: I think the most straightforward approach is to use L'Hôpital's rule:
$\begin{align}\lim\limits_{n\to\infty}1 - n\log\left(\dfrac{2n+1}{2n-1}\right) &= 1 - \lim\limits_{n\to\infty}\dfrac{\log\left(\dfrac{2n+1}{2n-1}\right)}{n^{-1}}\\
&= 1 - \lim\limits_{n\to\infty}\dfrac{\dfrac{2n-1}{2n+1}\dfrac{-4}{(2n-1)^2}}{-n^{-2}}\\
&= 1 - \lim\limits_{n\to\infty}\dfrac{4n^2}{(2n+1)(2n-1)}\\
&= 1-1\\
 &= 0
\end{align}$
L'Hôpital's rule is used in moving from the first line to the second, because the limit at the end of the first line is a $\frac00$ indeterminate form. The chain rule is applied in the numerator, with the derivative of the inside expression being taken by first rewriting it as $\dfrac{2n+1}{2n-1} = 1+\dfrac{2}{2n-1}$.
A: If you want to go beyond the limit itself, write $$1 - n \log \left(\dfrac{2n + 1}{2n - 1}\right)=1-n \log\left(1+ \frac{2}{2n - 1}\right)$$ Now, remembering that, for small values of $x$ $$\log(1+x)=x-\frac{x^2}{2}+O\left(x^3\right)$$ relace $x$ by $\frac{2}{2n - 1}$ which makes $$\log \left(\dfrac{2n + 1}{2n - 1}\right)=\log\left(1+ \frac{2}{2n - 1}\right)=\frac{2}{2n - 1}-\frac 12 \left(\frac{2}{2n - 1} \right)^2+\cdots$$ Now, use long division to get $$\log \left(\dfrac{2n + 1}{2n - 1}\right)=\frac{1}{n}+\frac{1}{12 n^3}+O\left(\frac{1}{n^4}\right)$$ So, for large values of $n$ $$1 - n \log \left(\dfrac{2n + 1}{2n - 1}\right)=-\frac{1}{12 n^2}+O\left(\frac{1}{n^3}\right)$$ which shows the limit and also how it is approached.
