Evaluating $\lim_{n\to\infty}{n\left(\ln(n+2)-\ln n\right)}$ I am trying to find$$\lim_{n\to\infty}{n\left(\ln(n+2)-\ln n\right)}$$ 
But I can't figure out any good way to solve this. 
Is there a special theorem or method to solve such limits? 
 A: $$ n\left(\log(n+2)-\log(n)\right) = n \int_{n}^{n+2}\frac{dx}{x} = \int_{0}^{2}\frac{n}{n+x}\,dx $$
so, by the dominated convergence theorem,
$$ \lim_{n\to +\infty}n\left(\log(n+2)-\log(n)\right) = \int_{0}^{2}1\,dx = \color{red}{2}.$$
A: Why not elementary? $n(\ln(n+2)-\ln n)=\ln(\frac{n+2}{n})^n=\ln(1+\frac{2}{n})^n \to \ln e^2=2$
A: $n\left( \ln(n+2)-\ln(n) \right)=\frac{\ln(1+2/n)}{1/n}$  so the limit is 2 since it is the derivative of $\ln(1+2x)$ at 0.
A: Another approach: (less elementary, but systematic and generalizable.) If you know of (and can use) Taylor expansions:
 we will use that, when $u\to 0$, $$\ln (1+u) = u + o(u).$$
(In what follows, what will be "our $u$" is $\frac{2}{n}\xrightarrow[n\to\infty]{} 0$.)
You then have
$$
\ln(n+2) = \ln\left(n(1+\frac{2}{n})\right)  = \ln n + \ln\left(1+\frac{2}{n}\right) = \ln n + \frac{2}{n} + o\left(\frac{1}{n}\right)
$$
so that
$$
n\left(\ln(n+2) - \ln n\right) = n\left(\frac{2}{n} + o\left(\frac{1}{n}\right)\right) = 2 + o(1) \xrightarrow[n\to\infty]{} 2.
$$
A: Another method: By the mean value theorem, we have
$$
\ln(n+2) - \ln(n) = \frac{1}{\xi_n} \cdot [(n+2)-n] = \frac{2}{\xi_n}
$$
for some $\xi_n \in (n, n+2)$.
Thus, $1 \leq \frac{n}{\xi_n} \leq \frac{n}{n+2} \to 1$ as $n\to \infty$, which again implies that the limit is $2$.
A: I like l'Hospital Rule and we can define 
$$f(x)=x\left(\log(x+2)-\log x\right)=\frac{\log\left(1+\frac2x\right)}{\frac1x}$$
If we do $\;x\to\infty\;$ then we get above an ideterminate $\;\frac00\;$ , so:
$$\lim_{x\to\infty} f(x)\stackrel{l'H}=\lim_{x\to\infty}\frac{-\frac2{x^2}\cdot\frac x{x+2}}{-\frac1{x^2}}=\lim_{x\to\infty}2\,\frac x{x+2}=2$$
As the function's limit is two, then also $\;\lim\limits_{n\to\infty}f(c_n)=2\;$ for any sequence such that $\;\lim\limits_{n\to\infty}c_n=\infty$ (I think this is called Heine's sequential definition of limit)
A: Hint
$$\ln(n+2)-\ln(n) = \ln\bigg (\frac{n+2}{n}\bigg)$$
