1. We need to evaluate $L=\lim_{x\to 0}\frac{\int_0^{x^2} \sin\sqrt{t}\,dt}{x^3}.$

2. Use the substitution $u=\sqrt{t}\quad \Rightarrow \quad t=u^2,\; dt=2u\,du.$ When $t=0$, $u=0$, and when $t=x^2$, $u=\sqrt{x^2}=|x|$.

So the integral becomes $\int_0^{x^2} \sin\sqrt{t}\,dt = \int_0^{|x|} \sin u\,(2u)\,du = 2\int_0^{|x|} u\sin u\,du.$ Hence $L=\lim_{x\to 0} \frac{2\int_0^{|x|} u\sin u\,du}{x^3}.$

3. For small $u$, $\sin u \sim u,$ so $u\sin u \sim u^2.$ Therefore, $2\int_0^{|x|} u\sin u\,du \sim 2\int_0^{|x|} u^2\,du = 2\cdot \frac{|x|^3}{3} = \frac{2|x|^3}{3}.$

Thus, $L \sim \frac{\frac{2|x|^3}{3}}{x^3} = \frac{2}{3}\cdot \frac{|x|^3}{x^3}.$

4. Now observe:

• if $x\to 0^+$, then $|x|^3=x^3$, so the limit is $\frac{2}{3}$;
• if $x\to 0^-$, then $|x|^3=(-x)^3=-x^3$, so $\frac{|x|^3}{x^3}=-1,$ and the limit is $-\frac{2}{3}$.

So the two-sided limit does not exist.

5. We can also verify exactly by integrating by parts: $\int u\sin u\,du = -u\cos u + \sin u.$ Thus, $2\int_0^{|x|} u\sin u\,du = 2\big(-|x|\cos|x|+\sin|x|\big).$ Hence $L=\lim_{x\to 0}\frac{2\big(\sin|x|-|x|\cos|x|\big)}{x^3}.$ Using $\sin y = y-\frac{y^3}{6}+o(y^3),\qquad \cos y=1-\frac{y^2}{2}+o(y^2),$ we get