Answer
$$1$$
Work Step by Step
The Taylor series for $\ln (1+x)$ can be defined as: $\ln (1+x)=x-\dfrac{ x^2}{2}+\dfrac{x^3}{3}-....$ and the Taylor series for $\sin x $ can be defined as: $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-....$
We have
$\lim\limits_{x \to 0} \dfrac{\ln (1+x^3)}{x \sin x^2}=\lim\limits_{x \to 0} \dfrac{x^3-\dfrac{x^6}{2}+\dfrac{x^9}{3} -....}{x^3-\dfrac{x^7}{6}+\dfrac{x^{11}}{120} -} \\= \dfrac{\lim\limits_{x \to 0}[x^3-\dfrac{x^6}{2}+\dfrac{x^9}{3} -....]}{\lim\limits_{x \to 0}[x^3-\dfrac{x^7}{6}+\dfrac{x^{11}}{120} -...]} \\ =\dfrac{1-0+0-...}{1-0+0-.....}\\=1$
