Limit of a sine-integral expression

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.