The problem I am stuck on asks the reader to find the following limit: $$\lim_{n \rightarrow \infty} \int^{n}_{0} \left(1+\frac{x}{n}\right)^{-n} \log\left(2+ \cos\left(\frac{x}{n}\right)\right)\ \mathrm dx.$$ The section I am working on contains all your basic limit theorem in measure theory (Monotone Covergence Theorem, Fatou's Lemma, Dominated Convergence Theorem). I know I am probably overseeing an application of one of them. Help would be greatly appreciated.

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    $\begingroup$ Domainated convergence theorem is your tool. $\endgroup$ – Mhenni Benghorbal Jul 18 '14 at 22:43

For $0 \leq y \leq 1$ we have $\log(1+y) \geq y - y^2/2$, so for $0 < x \leq n$ we have

$$ \begin{align} \left(1+\frac{x}{n}\right)^{-n} &= \exp\left[ -n \log\left(1+\frac{x}{n}\right)\right] \\ &\leq \exp\left[-n \left(\frac{x}{n} - \frac{x^2}{2n^2}\right)\right] \\ &= \exp\left[-x + \frac{x^2}{2n}\right] \\ &\leq \exp\left[-x + \frac{x^2}{2x}\right] \\ &= e^{-x/2}. \end{align} $$

Now apply dominated convergence.



$$\ln(2+\cos(x/n)) \leq \ln(3).$$

  • $\begingroup$ For the second term $\log (2+...) $ you may use the inequality $$\frac {x}{1+x} < \log (1+x)<x $$ valid for all, $ x>-1$ with $ x\ne 0$. $\endgroup$ – mwomath Jul 19 '14 at 3:11

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