How to integrate this?
Evaluate $$\lim_{x\to0}\frac{\displaystyle\int_0^x(x-t)\sin t^2\ dt}{x\sin^3x}.$$
I've had difficulty in using L'Hopital rule. At the same time failed to really understand how to differentiate or evaluate the limit of the $x\sin^3x$ and $\int_0^x(x-t)\sin t^2\ dt$
Would like to appreciate your help
Answer
Use repeatedly $\sin u=u\,{\rm sinc}(u)$, whereby $\lim_{u\to0}{\rm sinc}(u)={\rm sinc}(0)=1$. We have $$\int_0^x (x-t)\sin(t^2)\>dt=x^4\int_0^1(1-\tau)\,\tau^2{\rm sinc}(x^2\tau^2)\>d\tau$$ and $$\>x\sin^3 x=x^4\>{\rm sinc}^3(x)\ ,$$ so that $${\int_0^x (x-t)\sin(t^2)\>dt \over x\sin^3 x}={\int_0^1(1-\tau)\,\tau^2\bigl(1+r(x,\tau)\bigr)\>d\tau\over 1+\bar r(x)}\ ,$$ whereby $\lim_{x\to0}r(x,\tau)=0$ uniformly in $\tau$, and $\lim_{x\to0}\bar r(x)=0$ as well. It follows that the limit in question is $$\int_0^1(1-\tau)\,\tau^2\>d\tau={1\over12}\ .$$
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