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.
 A: Edit: The second half of this is nonsense. See the comments below...
Say the integrand is $f$. If $0<x\le1$ then $|f(x)|\le 1$, while if $x\ge1$ then $|f(x)|\le 1/x^2$, since $|\sin(t)|\le|t|$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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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}
A: By the self-adjointness of the Laplace transform
$$\int_{0}^{+\infty}\frac{\sin x}{x}\cdot \frac{1}{nx+\frac{1}{n}}\,dx =\frac{1}{n} \int_{0}^{+\infty}\left(\frac{\pi}{2}-\arctan(s)\right)e^{-s/n^2}\,ds $$
where the RHS is more manageable than the LHS since no oscillating functions are involved.
We have
$$ \int_{0}^{n^3}\left(\frac{\pi}{2}-\arctan(s)\right)\,ds =n^3\arctan\frac{1}{n^3}+\frac{1}{2}\log(1+n^6)=O(\log n)$$
$$ \int_{n^3}^{+\infty} e^{-s/n^2}\,ds = n^2 e^{-n}=O(1)$$
hence the wanted limit is clearly zero.
