Shorter Way to Solve $\lim_{x\to 0} \frac{x^3-\ln^3(1+x)}{\sin^2x-x^2}$ The limit to find is
$$\lim_{x\to 0} \frac{x^3-\ln^3(1+x)}{\sin^2x-x^2}$$
What I've tried was using the factorisation for $a^3-b^3$ and $a^2-b^2$ like so:
$$\lim_{x\to 0}\frac{(x-\ln(1+x))(x^2+x\ln(x+1)+\ln^2(x+1))}{(\sin x-x)(\sin x+x)}$$
but I don't know how to make a product of this so I get something meaningful (like a standard limit or something).
Lastly I've tried l'Hospital's rule and it works in the end, but after the second time the numerator and denominator are pretty long. Is there maybe a shorter solution to this?

Using $\sin(x)=x-\frac{x^3}{6}+\mathcal{O}(x^5)$ and $\ln(1+x)=x-\frac{x^2}{2}+\mathcal{O}(x^3)$:
$$\begin{aligned}\frac{x^3-\ln^3(1+x)}{\sin^2(x)-x^2}
&=\frac{x^3-\left(x-\frac{x^2}{2}+\mathcal{O}(x^3)\right)^3}{\left(x-\frac{x^3}{6}+\mathcal{O}(x^5)\right)^2-x^2}\\
&=\frac{x^3-\left(x^3- \frac{3 x^4}{2}+\mathcal{O}(x^5)\right)}{\left(x^2-\frac{x^4}{3}+\mathcal{O}(x^6)\right)-x^2}\\
&=\frac{\frac{3 x^4}{2}+\mathcal{O}(x^5)}{-\frac{x^4}{3}+\mathcal{O}(x^6)}
\end{aligned}$$
