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Prove:

$$\lim_{\varepsilon \rightarrow 0^+} \int_0^\infty \frac{\varepsilon}{\varepsilon +x} \sin(\frac{1}{x}) = 0$$

I'm reviewing a past exam question. Could anyone point me in the right direction for starting this problem? I've tried simplifying the integral by integration by parts or u-substitution, but that hasn't worked for me.

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  • $\begingroup$ Substitute $x\mapsto1/x$ and integrate by parts once. $\endgroup$ – Simply Beautiful Art Jul 28 at 2:19
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    $\begingroup$ A hint that may help you. Whenever $x>a>0$ (some $a$), you can just substitute $\epsilon$ and get 0 for that part. So for $\int_{a}^{\infty}$ you have guaranteed 0. Consider the remaining part, by analyzing the behavior of fraction. You can change it a little bit. $\endgroup$ – kolobokish Jul 28 at 2:19
  • $\begingroup$ Ok. One more hint. You should show boundedness of $\int_{0}^{a}$ connected with $a$. $\endgroup$ – kolobokish Jul 28 at 2:41
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Using $y=\frac{1}{x}$, $$\lim_{\varepsilon\to0^+}\int_0^\infty\frac{\sin\frac{1}{x}}{\varepsilon+x}dx=\lim_{\varepsilon\to0^+}\int_0^\infty\frac{1}{1+\varepsilon y}\frac{\sin y}{y}dy=\int_0^\infty\frac{\sin y}{y}dy=\frac{\pi}{2},$$where the penultimate $=$ uses the dominated convergence theorem. Thus$$\lim_{\varepsilon\to0^+}\int_0^\infty\frac{\varepsilon\sin\frac{1}{x}}{\varepsilon+x}dx=0.$$

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The integral from $0$ to $1$ tends to $0$ by DCT because $\sin \, t$ is bounded and $ \frac {\epsilon} {\epsilon+x} \leq 1$. For the integral from $1$ to $\infty$ use the fact that $|\sin \, t| \leq t$. So we only have to show that $\int_1^{\infty} \frac {\epsilon} {x(\epsilon+x)} \, dx \to 0$. Here the integrand is bounded by $\frac 1 {x^{2}}$ which is integrable, so DCT applies again.

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