How find this limit $\lim\limits_{n\to\infty}\int_{0}^{2\pi}\left(\frac{\sin{(nx)}}{x^2+n^2}+\frac{\cos{(nx)}}{1+x}\right)dx $ Find this limit
$$\lim_{n\to\infty}\int_{0}^{2\pi}\left(\dfrac{\sin{(nx)}}{x^2+n^2}+\dfrac{\cos{(nx)}}{1+x}\right)dx $$
Maybe use Division of the integral?Thank you
 A: You can easily prove that $\dfrac{\sin(nx)}{x^{2}+n^{2}} \rightarrow 0$ uniformly for $x \in [0,2\pi]$, which justifies the swap between the limit and the integral, hence $$\lim_{n\rightarrow \infty} \int_{0}^{2\pi}{\frac{\sin(nx)}{x^{2}+n^{2}}dx}= \int_{0}^{2\pi}{0 \,dx}=0$$ For the second one a simple integration by parts followed by an estimation yields $$\left|\int_{0}^{2\pi}{\frac{\cos(nx)}{1+x}dx}
\right|\leq\frac{1}{n}\int_{0}^{2\pi}{\frac{dx}{|1+x|^{2}}}=\frac{1}{n}\cdot\left|\frac{2\pi}{1+2\pi}\right|\rightarrow 0$$ as $n \rightarrow \infty.$
A: For the 1st term inside the integral, we have: /sin(nx)/(x^2 + n^2)/ < 1/n^2. So by taking limit the integral of this term is zero. For the 2nd term, using integration by part:
Int(0 --> 2*pi)cos(nx)/(1 + x)dx = Int(0 --> 2*pi)1/(n(x + 1))*d(sin(nx)) = 0 - Int(0 --> 2*pi)sin(nx)/(n(x + 1)^2)dx. Observe that /sin(nx)/(n(x + 1)^2/ < 1/n. So taking limit this integral is zero. And the answer is: 0.
A: We know that $$\dfrac{\sin nx}{x^2+n^2}\to 0$$ uniformly. So the first part tends to 0. Now consider $$\int_0^{2\pi}\dfrac{\cos nx}{1+x}dx=\int_0^{\pi/n}\dfrac{\cos nx}{1+x}dx+\int_{\pi/n}^{2\pi/n}\dfrac{\cos nx}{1+x}dx+\dots=\int_0^{\pi/n}{\cos nx}\dfrac 1{1+x}-\dfrac{1}{1+\pi/n+x}dx+\int_{2\pi/n}^{3\pi/n}{\cos nx}\dfrac 1{1+x}-\dfrac{1}{1+\pi/n+x}dx+\int_{4\pi/n}^{5\pi/n}{\cos nx}\dfrac 1{1+x}-\dfrac{1}{1+\pi/n+x}dx+\dots\leq \pi^2/n$$. (The inequality follows as we are adding less than n terms which are all less than $\pi^/n$ as the terms in the integral are less than $\pi/n$ and the intervals are of size $\pi /n$)So it tends to zero as it is also positive.
