Lim x→0^+(1)/(x^5(∫0^x^2sin √ t dt-2/3x^3 equals

Use the Maclaurin expansion $\sin u=u-\dfrac{u^3}{6}+\cdots$ with $u=\sqrt t$, giving $\sin\sqrt t=t^{1/2}-\dfrac{t^{3/2}}{6}+O(t^{5/2})$. Integrating term by term from $0$ to $x^2$: $\int_0^{x^2}\sin\sqrt t\,dt=\dfrac23 t^{3/2}-\dfrac{1}{6}\cdot\dfrac{2}{5}t^{5/2}+\cdots=\dfrac23x^{3}-\dfrac1{15}x^{5}+O(x^7)$. Subtracting $\tfrac23x^3$ leaves $-\tfrac{1}{15}x^5+O(x^7)$, and dividing by $x^5$ and letting $x\to0^+$ gives $-\tfrac1{15}$. The sign is negative because $\sin u<u$; a common slip is to forget the factor $\tfrac25$ from integrating $t^{3/2}$, which yields $-\tfrac1{30}$-type values.