<s>[INST] 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?
 [/INST] 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
$$\frac{x^3 - \ln^3(1+x) }{\sin^2(x) - x^2}  \simeq \frac{x^3 - x^3 +\frac{3}{2}x^4}{x^2 - \frac{x^4}{3} - x^2}  = \frac{\frac{3}{2}x^4}{- \frac{x^4}{3}}  = - \frac{9}{2}$$
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