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Find $$\lim_{n \to \infty} \int_{0}^n\frac{1}{n+n^2 \sin \left( \dfrac{x}{n^2} \right)} \mbox{d}x $$

I've defined $f_n(x) = \begin{cases} \dfrac{1}{n+n^2 \sin \left( \frac{x}{n^2} \right)}, & \text{if} \ \ x \in [0,n] \\ 0, & \text{if} \ x >n\end{cases}$

Of course $f_n \to 0$. I have to find $\displaystyle\lim_{n \to \infty} \int_{0}^\infty f_n(x) \mbox{d}x $, I'm trying to use Lebesgue theorem but i can't find function $g$ such that $|f_n(x)| \le g(x)$ and $\displaystyle\int _0 ^\infty g(x) \mbox{d}x < \infty$.

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marked as duplicate by Ayman Hourieh, Amzoti, Did, Micah, Chris Eagle Feb 26 '13 at 21:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

As I remember, exactly the same question with an excellent answer already exists in this community, though I do not remember the link... – Sangchul Lee Feb 26 '13 at 20:05

I would just rescale the integral: let $x=n u$, then the integral becomes

$$\lim_{n \rightarrow \infty} \int_0^1 \frac{du}{1+n \sin{\left ( \frac{u}{n}\right)}} = \int_0^1 \frac{du}{1+u} = \log{2}$$

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