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?
 A: Divide numerator and denominator by $x^4$ and then the denominator is handled as $$\frac{\sin x-x}{x^3}\cdot\frac{\sin x +x} {x} \to (-1/6)(2)=-\frac{1}{3}$$ The numerator on the other hand is handled as $$\frac{x-\log(1+x)}{x^2}\cdot\frac {x^2+x\log(1+x)+\log^2(1+x)} {x^2} \to\frac{1}{2}\cdot(1+1+1)=\frac{3}{2}$$ and hence the desired limit is $-9/2$.
A: Taylor Series around $0$ of $\ln(1+x)$:
$$\ln(1+x) = x - \frac{x^2}{2} + O(x^3)$$
$$\ln^3(1+x) = x^3 -\frac{3}{2}x^4 + O(x^5)$$
Taylor Series around $0$ of $\sin(x)$:
$$\sin(x) = x - \frac{x^3}{6} + O(x^5)$$
$$\sin^2(x) = x^2 - \frac{x^4}{3} + O(x^6)$$
So
$$\frac{x^3 - \ln^3(1+x) }{\sin^2(x) - x^2}  \simeq \frac{x^3 - x^3 +\frac{3}{2}x^4}{x^2 - \frac{x^4}{3} - x^2}  = \frac{\frac{3}{2}x^4}{- \frac{x^4}{3}}  = - \frac{9}{2}$$
A: Using $\sin(x)=x-\frac{x^3}{6}+\mathcal{O}(x^5)$ and $\ln(1+x)=x-\frac{x^2}{2}+\mathcal{O}(x^3)$:
$$\begin{aligned}\frac{x^3-\ln^3(1+x)}{\sin^2(x)-x^2}
&=\frac{x^3-\left(x-\frac{x^2}{2}+\mathcal{O}(x^3)\right)^3}{\left(x-\frac{x^3}{6}+\mathcal{O}(x^5)\right)^2-x^2}\\
&=\frac{x^3-\left(x^3- \frac{3 x^4}{2}+\mathcal{O}(x^5)\right)}{\left(x^2-\frac{x^4}{3}+\mathcal{O}(x^6)\right)-x^2}\\
&=\frac{\frac{3 x^4}{2}+\mathcal{O}(x^5)}{-\frac{x^4}{3}+\mathcal{O}(x^6)}
\end{aligned}$$
A: Hint: By a  series Expansion we get
$-\frac{9}{2}+\frac{21}{4}x-\frac{249}{40}x^2+\frac{13}{2}x^3-\frac{18619}{2800}x^4+\frac{33647}{5040}x^5+...$
A: You could use Taylor series at $x=0$ to compute the limit. Commonly used Taylor Series
$$\lim_{x\to 0} \frac{x^3-\ln^3(1+x)}{\sin^2x-x^2}$$
By using Taylor series we get
$$-\frac92+\frac{21x}{4}-\frac{249x^2}{40}+\frac{13x^3}{2}-.....O(x^6)$$at $x=0$ we get
$$\lim_{x\to 0} \frac{x^3-\ln^3(1+x)}{\sin^2x-x^2}=-\frac92$$
