Answer
$$1$$
Work Step by Step
Recall the Taylor series for $\ln (1+x)=x-\dfrac{ x^2}{2}+\dfrac{x^3}{3}-....$ and $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-....$
$\lim\limits_{x \to 0} \dfrac{\ln (1+x^3)}{x \sin x^2}= \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} -...]} $
or, $=\dfrac{1-0+0-...}{1-0+0-.....}$
or, $\lim\limits_{x \to 0} \dfrac{\ln (1+x^3)}{x \sin x^2}=1$
