Limit of a sine-integral expression How can I see that
\begin{align*}
\lim_{x\searrow0}\frac{\dfrac{\pi}{6}-\dfrac{1}{3}\displaystyle\int_0^{1/x^3}\dfrac{\sin t}{t}\,\mathrm{d}t}{x}=0?
\end{align*}
I tried L'Hôpital's rule (since both the numerator and the denominator tend to zero and both are differentiable for $x>0$), but it doesn't work, since the resulting expression possesses no limit. I'm pretty sure the answer is relatively simple, I just ran out of ideas.
Please share your views on this with me. Thank you.
 A: Since $\int_0^{+\infty}\frac{\sin t}t\mathrm dt=\frac{\pi}2$, the only task is to show that 
$$\lim_{x\downarrow 0}\frac 1x\int_{1/x^3}^{+\infty}\frac{\sin t}t\mathrm dt=0.$$
This can be done by an integration by parts (integrate $\sin t$, differentiate $\frac 1t$).
A: Write the numerator as $\int_{1/x^3}^{+\infty} \frac{\sin t}{t}dt$ and integrate by parts.
A: You can simply recall the (complex analysis) proof of 
$$\int_0^{\infty}\frac{\sin t}{t}dt=\frac{\pi}{2}$$
and estimate the remainder term 
$$\int_R^{\infty}\frac{\sin t}{t}dt$$
to see in what degree it is approaching zero when $R\to\infty$. Then the answer will be straightforward.
A: Thank you so much for your help. Based on your guidance, here's how I did it (it might be more complicated than necessary, but I have a rigorous proof style, so excuse me for the unnecessary confusion):
For a given $x>0$, fix $b>1/x^3$. Then,
\begin{align*}
\int_{1/x^3}^{b}\frac{\sin t}{t}\,\mathrm{d} t=\int_0^{b}\frac{\sin t}{t}\,\mathrm{d} t-\int_0^{1/x^3}\frac{\sin t}{t}\,\mathrm{d} t.
\end{align*}
Therefore,
\begin{align*}
\lim_{b\nearrow+\infty}\left(\int_{1/x^3}^{b}\frac{\sin t}{t}\,\mathrm{d} t\right)=\frac{\pi}{2}-\int_0^{1/x^3}\frac{\sin t}{t}\,\mathrm{d} t.
\end{align*}
Hence, the original limit in question can be rearranged as follows:
\begin{align*}
\lim_{x\searrow0}\left(\frac{1}{3x}\lim_{b\nearrow+\infty}\left(\int_{1/x^3}^{b}\frac{\sin t}{t}\,\mathrm{d} t\right)\right)=\lim_{x\searrow0}\left(\lim_{b\nearrow+\infty}\left(\frac{1}{3x}\underbrace{\int_{1/x^3}^{b}\frac{\sin t}{t}\,\mathrm{d} t}_{(\heartsuit)}\right)\right).
\end{align*}
Now, for a given $x>0$ and $b>1/x^3$, $(\heartsuit)$ can be expanded, indeed, using integration by parts:
\begin{align*}
\int_{1/x^3}^{b}\frac{\sin t}{t}\,\mathrm{d} t=-\frac{\cos t}{t}\Bigg|_{t=1/x^3}^{t=b}-\int_{1/x^3}^{b}\frac{\cos t}{t^2}\,\mathrm{d}t=-\frac{\cos b}{b}+x^3\cos(1/x^3)-\int_{1/x^3}^{b}\frac{\cos t}{t^2}\,\mathrm{d}t.
\end{align*}
Hence,
\begin{align*}
\lim_{x\searrow0}\left(\lim_{b\nearrow+\infty}\left(\frac{1}{3x}(\heartsuit)\right)\right)=\lim_{x\searrow0}\left(\frac{1}{3x}\left(-\underbrace{\lim_{b\nearrow+\infty}\left(\frac{\cos b}{b}\right)}_{=0}\right)\right)+\underbrace{\lim_{x\searrow0}\left(\frac{x^2}{3}\cos(1/x^3)\right)}_{=0}-\lim_{x\searrow0}\left(\underbrace{\lim_{b\nearrow+\infty}\left(\frac{1}{3x}\int_{1/x^3}^{b}\frac{\cos t}{t^2}\,\mathrm{d}t\right)}_{(\spadesuit)}\right).
\end{align*}
But note that
\begin{align*}
\lim_{x\searrow0}\,\left|\,(\spadesuit)\,\right|=\lim_{x\searrow0}\,\left|\,{\lim_{b\nearrow+\infty}\left(\frac{1}{3x}\int_{1/x^3}^{b}\frac{\cos t}{t^2}\,\mathrm{d}t\right)}\,\right|\underbrace{=}_{\text{$|\cdot|$ cuous}}
\lim_{x\searrow 0}\left(\lim_{b\nearrow+\infty}\left(\frac{1}{3x}\,\left|\,\int_{1/x^3}^{b}\frac{\cos t}{t^2}\,\mathrm{d}t\,\right|\right)\right)\leq\lim_{x\searrow 0}\left(\lim_{b\nearrow+\infty}\left(\frac{1}{3x}\int_{1/x^3}^{b}\frac{\left|\,\cos t\,\right|}{t^2}\,\mathrm{d}t\right)\right)\leq\lim_{x\searrow 0}\left(\lim_{b\nearrow+\infty}\left(\frac{1}{3x}\int_{1/x^3}^{b}\frac{\mathrm{d}t}{t^2}\right)\right)=\lim_{x\searrow 0}\left(\lim_{b\nearrow+\infty}\left(\frac{x^3-\dfrac{1}{b}}{3x}\right)\right)=\lim_{x\searrow0}\frac{x^2}{3}=0.
\end{align*}
Consequently, $\lim_{x\searrow0}\,\left|\,(\spadesuit)\,\right|=0$, implying that $\lim_{x\searrow0}(\spadesuit)=0$, too, completing the proof.
Thanks to everyone for the hints once again! @njguliyev @DavideGiraudo @shallpion
