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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?

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Why not elementary? $n(\ln(n+2)-\ln n)=\ln(\frac{n+2}{n})^n=\ln(1+\frac{2}{n})^n \to \ln e^2=2$

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$$ n\left(\log(n+2)-\log(n)\right) = n \int_{n}^{n+2}\frac{dx}{x} = \int_{0}^{2}\frac{n}{n+x}\,dx $$ so, by the dominated convergence theorem, $$ \lim_{n\to +\infty}n\left(\log(n+2)-\log(n)\right) = \int_{0}^{2}1\,dx = \color{red}{2}.$$

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    $\begingroup$ Weeeeeeeeew, I like it. So many different methods for one limit :) $\endgroup$ – PhoemueX Jan 31 '16 at 21:33
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$n\left( \ln(n+2)-\ln(n) \right)=\frac{\ln(1+2/n)}{1/n}$ so the limit is 2 since it is the derivative of $\ln(1+2x)$ at 0.

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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 $$ n\left(\ln(n+2) - \ln n\right) = n\left(\frac{2}{n} + o\left(\frac{1}{n}\right)\right) = 2 + o(1) \xrightarrow[n\to\infty]{} 2. $$

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    $\begingroup$ Nice explanation. (+1). $\endgroup$ – user98186 Jan 31 '16 at 21:41
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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 $2$.

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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)

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Hint

$$\ln(n+2)-\ln(n) = \ln\bigg (\frac{n+2}{n}\bigg)$$

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