Let $\displaystyle F(x) = \int_0^{x^2} \sin\sqrt t\,dt \qquad \text{for every real } x.$ The upper limit $x^2$ is the same for $x$ and for $-x$, so $F$ is an even function. Near $t = 0$ the integrand behaves like $\sqrt t$, and that decides how fast $F(x)$ goes to $0$. Recall that $\sqrt{x^2} = |x|$, not $x$, which matters as soon as $x$ is negative. Two of the statements are about what happens very close to $x = 0$, from one side or from both sides, and one is about the sign of $F$ far from $0$. The substitution $t = w^2$ gives $F$ in closed form, but most of the statements can be decided without it. Which of the following statements are true?