Calculate $\lim_{n\to \infty} \int_0^1 {nx\sin(x)\over n^3x^2 +1}{\rm d}x$ I want to calculate $\lim_{n\to \infty} \int_{[0,1]} {nx\sin(x)\over n^3x^2 +1}{\rm d}x$
I started with $f_n={nx\sin(x)\over n^3x^2 +1}\lt {nx\sin(x)\over n^3x^2 }={\sin(x)\over n^2x}$
How can I do next? 
 A: Define $f_n : [0,1] \to \mathbb{R}$ with $f_n(x) = \frac{nx\sin(x)}{n^3x^2 +1}$.
We have $\lim_{n\to\infty} f_n(x) = 0$ for all $x \in [0,1]$ because
$$|f_n(x)| = \frac{nx\left|\sin(x)\right|}{n^3x^2 +1} \le \frac{n}{n^3x^2+1} = \frac1{n^2x^2+\frac1n} \xrightarrow{n\to\infty} 0$$
Also, we have
$$|f_n(x)| = \frac{nx\left|\sin(x)\right|}{n^3x^2 +1} \le \frac{nx^2}{n^3x^2+1} \le \frac1{n^2}\cdot \frac{nx^2}{nx^2+1} \le \frac1{n^2} \le 1$$
with $$\int_{[0,1]} 1\,d\lambda(x) = 1< +\infty$$
so the sequence $(f_n)_n$ is dominated by an integrable function.
The Lebesgue dominated convergence theorem gives
$$\lim_{n\to\infty} \int_{[0,1]} f_n\,d\lambda = \int_{[0,1]}\Big(\lim_{n\to\infty} f_n\Big) \,d\lambda = \int_{[0,1]} 0\,d\lambda = 0$$
A: Hint :Try to use Dominated convergence Theorem.
Then change the integral and the Limit and use the fact that
$$lim_{n \rightarrow\infty} \frac{nx\  sin x}{n^3x^2+1}=0$$
Therefore
$$lim_{n \rightarrow \infty} \int_{0}^{1}\frac{nx\ sin x}{n^3x^2+1}=\int_{0}^{1}lim_{n \rightarrow \infty}\frac{nx\ sin x}{n^3x^2+1}=0$$
