I use the fundemental theorem of calculus
$$
\displaystyle\lim_{x\to 0}\frac{\displaystyle\int_0^{x^2}\sin{(\sqrt t)}dt}{x^3}=\frac{F_{(x^2)}-F_{(0)}}{x^3}="\frac{0}{0}"
$$
Than apply L'hopital rule
$$
\displaystyle\lim_{x\to 0}\frac{f_{(x^2)}-f_{(0)}}{3x^2}=\displaystyle\lim_{x\to 0}\frac{\sin(\sqrt x^2)-\sin(0)}{3x^2}=\displaystyle\lim_{x\to 0}\frac{\sin(x)}{3x^2}="\frac{0}{0}"
$$
Than L'hopital again:
$$
\displaystyle\lim_{x\to 0}\frac{\cos(x)}{6x}
$$
This limit does not exist.
Best Answer
When differentiating your integral, you forget to apply the chain rule, $$ \frac{d}{dx}\int_0^{x^2}\sin(\sqrt{t})\,dt=\sin(\sqrt{x^2})\cdot\frac{d}{dx}x^2=\sin (|x|)\cdot 2x. $$ (In the end, you will conclude that the limit does not exist.)