Evaluate $$L=\lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}} \sin \sqrt{t} d t}{x^{3}}$$
Since the limit is $\frac{0}{0}$ form, Using L'Hopital's rule and Leibniz rule we get $$L=\lim_{x \to 0}\frac{2x\sin |x|}{3x^2}$$ $\implies$ $$L=\frac{2}{3}\lim_{x \to 0}\frac{\sin|x|}{x}$$ Now we know that limit exists only when right and left hand limits are finite and equal. $$L^+=\lim_{h \to 0^+}\frac{\sin h}{h}=1$$ $$L^-=\lim_{h \to 0^+}\frac{\sin|-h|}{-h}=-1$$ Since $$L^+ \ne L^-$$ we can say limit Does not exists. But book answer is $\frac{2}{3}$
Indisputably, the numerator is an even function and the denominator an odd one. As they both turn out to be $\Theta(x^3)$, the left and right limits are finite and indeed differ in sign (as you say, $\pm\frac23$).
You can also address the question with
$$\int_0^{x^2}\sin\sqrt t\,dt\sim\int_0^{x^2}\sqrt t\,dt=\frac23(x^2)^{3/2}=\frac23|x|^3.$$