$$\lim_{n \to{+}\infty}{\int_{0}^{\infty}}\frac{n\sin\frac{x}{n}}{x(1+x^2)}dx$$
$$\lim_{n \to{+}\infty}{\int_{\mathbb{R^2}}}\frac{\arctan(n\left\|{x}\right\|)}{(1+\left\|{x}\right\|^2)^2}dx$$
Could anyone to help me in this two integrals? Thank you very much!
On
Following Integrand's comment and Oliver Diaz' suggestion for the second limit, it follows that $$ \begin{align} \lim_{n\rightarrow\infty}\int_{\mathbb{R}^2}\frac{\arctan(n\|x\|)}{(1+\|x\|^2)^2}\,dx&=\frac{\pi}{2}\int_{\mathbb{R}^2}\frac{dx}{(1+\|x\|^2)^2}\\ &=\frac{\pi}{2}\int^\infty_0\int^{2\pi}_0\frac{r}{(1+r^2)^2}\,d\theta\,dr\\ &=\pi^2\int^\infty_0\frac{r\,dr}{(1+r^2)^2}=\frac{\pi^2}{2}\int^\infty_0\frac{2r\,dr}{(1+r^2)^2}\\ &=-\frac{\pi^2}{2}\frac{1}{1+r^2}\Big|^\infty_0=\frac{\pi^2}{2} \end{align} $$
Hint: Dominated integration
For the first integral for instance using the fact that $|\sin y|/|y|\leq 1$ for all $y$ (with the convention is $\frac{\sin y}{y}=1$ for $y=0$), one gets
$$ \frac{n\sin\tfrac{x}{n}}{x(1+x^2)}\leq \frac{1}{1+x^2}$$
As $\lim_{n\rightarrow\infty}\frac{n\sin\tfrac{x}{n}}{x}=1$ for each $x$, by dominated convergence
$\lim_{n\rightarrow\infty}\int^\infty_0\frac{n\sin\tfrac{x}{n}}{x(1+x^2)}\,dx =\int^\infty_0\frac{1}{1+x^2}\,dx=\arctan(x)|^\infty_0=\frac{\pi}{2}$
For the second integral, notice that $|\arctan(y)|\leq\frac{\pi}{2}$ and that $\lim_{y\rightarrow\infty}\arctan(y)=\frac{\pi}{2}$. I leave the remaining details to you.