How do I show $\lim_{n \to \infty} \int_0^\infty \frac{n}{n^2+x}\sin(\frac{1}{x})\, dx = 0\,$? How do I show $$\lim_{n \to \infty} \int_0^\infty \frac{n}{n^2+x}\sin\left(\frac{1}{x}\right)\, dx = 0\,\,?$$ I've tried splitting into the cases where $x \leq 1$ and $x \geq 1$ but I am having trouble finding bounds so that I can apply the dominated convergence theorem.

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\begin{align}
&\bbox[15px,#ffd]{\lim_{n \to \infty}\int_{0}^{\infty}
{n \over n^{2} + x}\,\sin\pars{1 \over x}\,\dd x}
\,\,\,\stackrel{x\ \mapsto\ 1/x}{=}\,\,\,
\lim_{n \to \infty}\bracks{{1 \over n}\int_{0}^{\infty}
{\sin\pars{x} \over \pars{x + 1/n^{2}}x}\,\dd x}
\\[5mm] = &\
\lim_{n \to \infty}\bracks{{\pi n \over 2} -
n\int_{0}^{\infty}{\sin\pars{x} \over x + 1/n^{2}}\,\dd x}
\\[5mm] = &\
\lim_{n \to \infty}\pars{{\pi n \over 2} -
n\braces{\operatorname{Ci}\pars{1 \over n^{2}}\sin\pars{1 \over n^{2}} +
{1 \over 2}\bracks{\pi -2\operatorname{Si}\pars{1 \over n^{2}}}\cos\pars{1 \over n^{2}}}}
\end{align}
$\ds{\operatorname{Ci}}$ and
$\ds{\operatorname{Si}}$ are the
Cosine and Sine Integrals Functions, respectively.
As $\ds{z \to 0}$, $\ds{\quad\operatorname{Ci}\pars{z} \sim \gamma + \ln\pars{z} - {1 \over 4}\,z^{2}\quad}$ and
$\ds{\quad\operatorname{Si}\pars{z} \sim z - {1 \over 18}\,z^{3}\quad}$
from this link.

Therefore,
\begin{align}
&\bbox[15px,#ffd]{\lim_{n \to \infty}\int_{0}^{\infty}
{n \over n^{2} + x}\,\sin\pars{1 \over x}\,\dd x} =
-\lim_{n \to \infty}{\gamma - 2\ln\pars{n} \over n}
\\[5mm] = &\
-\lim_{n \to \infty}{\bracks{\gamma - 2\ln\pars{n + 1}} -
\bracks{\gamma - 2\ln\pars{n}} \over \pars{n + 1} - n} =
2\lim_{n \to \infty}\ln\pars{1 + {1 \over n}} =
\bbox[15px,#ffd,border:1px solid navy]{0}
\end{align}
