$$ \lim_{x \to 0+}\frac{4e^{x^2}-4\cos(x^3)-2\sin(x^4)-8\sqrt{1+x^2}-x^4+8}{1-\sqrt[3]{1+13x^6}}$$ I used the property $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$ to get rid of the cubic root in the denominator and then use the Taylor expansions of order 3 centered around 0 of each term to simplify the limit, but I am left with $$ \lim_{x\rightarrow 0+}{\frac{(O(x^4)+2x^6+O(x^{11})-2x^4+\frac{x^{12}}{3}-\frac{x^{20}}{60}-O(x^4)-O(x^4)-x^2)(3+13x^6+O(x^9)))}{-13x^6}}$$ but I can't factor anything to get rid of the $x^{6}$ below. I don't even know if what I did is correct. How should I proceed?
Note: I can´t use L´Hopital
Do not try to simplify and compose Taylor series.
First of all, consider the denominator; the binomial expansion or Taylor series give $${1-\sqrt[3]{1+13x^6}}=-\frac{13 }{3}x^6+O\left(x^{12}\right)$$ as you already noticed. This means that we need to develop each term of the numerator at least up to $O\left(x^{7}\right)$.
Now each term in numerator is simple. In order to get more that the limit itself, developing each term up to $O\left(x^{8}\right)$ will give
$$4 e^{x^2}=4+4 x^2+2 x^4+\frac{2 x^6}{3}+O\left(x^8\right)$$ $$-4\cos(x^3)=-4+2 x^6+O\left(x^8\right)$$ $$-2\sin(x^4)=-2 x^4+O\left(x^8\right)$$ $$-8\sqrt{1+x^2}=-8-4 x^2+x^4-\frac{x^6}{2}+O\left(x^8\right)$$ which make the numerator to be $$\frac{13 x^6}{6}+O\left(x^8\right)$$ So, we have the limit.