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

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.

• Substitute $x\mapsto1/x$ and integrate by parts once. – Simply Beautiful Art Jul 28 at 2:19
• 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. – kolobokish Jul 28 at 2:19
• Ok. One more hint. You should show boundedness of $\int_{0}^{a}$ connected with $a$. – kolobokish Jul 28 at 2:41

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$$\lim_{\varepsilon\to0^+}\int_0^\infty\frac{\varepsilon\sin\frac{1}{x}}{\varepsilon+x}dx=0.$$
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.