Evaluate $\lim_{x\to 0}\frac{1}{x^4}\int_{-x}^{0}\sin(t^3)\,dt$

Question

Evaluate $$\lim_{x\to 0}\frac{1}{x^4}\int_{-x}^{0}\sin(t^3)\,dt.$$

I use L'Hopital's Rule and get -1/4. The solution says 1/4. Any ideas?

Answer 1

You are right. Since in a neighbourhood of zero: $$\sin(t^3) = t^3+o(t^4)$$ we have: $$\int_{-x}^{0}\sin(t^3)\,dt = \int_{-x}^{0}t^3 dt + o\left(\int_{-x}^{0}t^4\,dt\right)$$ hence: $$\frac{1}{x^4}\int_{-x}^{0}\sin(t^3)\,dt=-\frac{1}{4}+o(x)$$ as $x\to 0$.

Answer 2

$\lim \limits_{x \to 0}\dfrac{\int \limits _{0}^{-x} \sin(t^3)dt}{ x^4} \stackrel{LH}{=} \lim \limits_{x \to 0}\dfrac{ \sin((-x)^3)(-x)'}{ 4x^3} = \lim \limits_{x \to 0}\dfrac{ \sin(-x^3)(-1)}{ 4x^3} = \lim \limits_{x \to 0}\dfrac{ \sin(x^3)}{ 4x^3} = \dfrac{1}{4}$