Calculate $$\lim_{n\to\infty}\int_0^n\frac{\mathrm{d}x}{n+n^2\sin\frac{x}{n^2}}$$, well first of all the function isn't monotonic, so I cant use the Lebesgue therem to go with the limit under the integral. What I could do is to express $sinx=x+O(x^3)$ then $$\lim_{n\to\infty}\int_0^n\frac{\mathrm{d}x}{n+n^2\sin\frac{x}{n^2}}=\lim_{n\to\infty}\int_0^n\frac{\mathrm{d}x}{n+x}=\lim_{n\to\infty} \ln2n-\ln n=\ln2$$, but i am not sure if this is correct and also would like to know why this is correct, as of course I ate the $O(x^3)$ when approximating sin function.

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    $\begingroup$ Double limit always needs extra care. Although your approach is totally justifiable, it would be easier to first substitute $x=nu$ and the pass limit to the inside of the integral. $\endgroup$ – Sangchul Lee Mar 20 '19 at 18:00
  • $\begingroup$ nice suggestion! $\endgroup$ – Ryszard Eggink Mar 20 '19 at 18:02


Since $u-{u^3\over 6}\le \sin u\le u$ for $0\le u<<1$ (small enough $u$) you can write $${1\over n+x}\le\frac{1}{n+n^2\sin\frac{x}{n^2}}\le \frac{1}{n+x-{x^3\over n^4}}\le{1\over n+x-{1\over n}}$$and integrate the sides.


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