# Calculate $\lim_{n\to\infty}\int_0^n\frac{\mathrm{d}x}{n+n^2\sin\frac{x}{n^2}}$

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.

• 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. – Sangchul Lee Mar 20 '19 at 18:00
• nice suggestion! – 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.