<s>[INST] Compute $\lim_{n\to\infty}\int_0^1\frac{x\sin{nx}}{1+x^2n^6}dx$ $\lim_{n\to\infty}\int_0^1\frac{x\sin{nx}}{1+x^2n^6}dx$ 
My Work:
By the comparison principle: $$\lim_{n\to\infty}\int_0^1\frac{x\sin{nx}}{1+x^2n^6}dx\le\lim_{n\to\infty}\int_0^1\frac x{1+x^2n^6}dx$$ I think that the way to compute this integral has to do with uniform convergence, but I cannot figure how to determine it.
 [/INST] Hint: For $n\ge1$, $$\left|\frac{x\sin(nx)}{1+x^2n^6}\right|\le x$$ so Dominated Convergence should apply if the domain of integration is $[0,1]$.

If the Domain of Integration is $\boldsymbol{[0,\infty)}$
Integration by Parts gives
$$
\begin{align}
\int_0^\infty\frac{x\sin(nx)}{1+x^2n^6}\,\mathrm{d}x
&=-\frac1n\int_0^\infty\frac{x}{1+x^2n^6}\,\mathrm{d}\cos(nx)\\
&=\frac1n\int_0^\infty\frac{1-x^2n^6}{\left(1+x^2n^6\right)^2}\cos(nx)\,\mathrm{d}x\\
&=\frac1{n^4}\int_0^\infty\frac{1-x^2}{\left(1+x^2\right)^2}\cos\left(\frac{x}{n^2}\right)\,\mathrm{d}x
\end{align}
$$
Now Dominated Convergence applies since
$$
\begin{align}
\int_0^\infty\left|\frac{1-x^2}{\left(1+x^2\right)^2}\right|\,\mathrm{d}x
&\le\int_0^\infty\frac{1+x^2}{\left(1+x^2\right)^2}\,\mathrm{d}x\\
&=\int_0^\infty\frac1{1+x^2}\,\mathrm{d}x\\
&=\frac\pi2
\end{align}
$$
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