<s>[INST] Show $\lim_{a \rightarrow \infty} \int_0^1 {f(x)x\sin(ax^2)}=0$. 
Suppose $f$ is integrable on $(0,1)$, then show $$\lim_{a \rightarrow \infty} \int_0^1 {f(x)x\sin(ax^2)}=0.$$

I tried to write $$(0,1) = \bigcup _{k=0}^{{a-1}} \left(\sqrt{\frac{k}{a}},\sqrt{\frac{k+1}{a}}\right),$$ but cannot make the integral converge to $0$.
 [/INST] Suggestion:
Split $[0, 1]$
into the intervals where
$ax^2 = 2\pi n$,
$ax^2 = \pi(2 n+1)$,
$ax^2 = \pi(2 n+2)$.
These are
$I_{2n}
=[\sqrt{\frac{2\pi n}{a}}, \sqrt{\frac{\pi(2 n+1)}{a}})
$
and
$I_{2n+1}
=[\sqrt{\frac{\pi(2 n+1)}{a}}, \sqrt{\frac{\pi(2 n+2)}{a}})
$.
Since
$\sin(ax^2) > 0$
in
$I_{2n}$
and
$\sin(ax^2) < 0$
in
$I_{2n+1}$,
show that
the integral over
$I_{2n} \cup I_{2n+1}$
goes to zero.
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