Find: $\lim_{x\to0}\frac{\ln(1+x^2)-x^2+\frac{x^4}{2}}{\sin x^6}$ How do I even approach such a question? Obviously l'Hospital rule would be a mess and I don't know how to even start. Most of the limits we deal with are way simpler than this.

L'Hopital:
$$\lim\limits_{x\to 0}\frac{-2x+2x^3+\frac{2x}{x^2+1}}{6x^5\cos(x^6)}=\lim\limits_{x\to 0}\frac{\frac{2x^5}{x^2+1}}{6x^5\cos(x^6)}=\lim\limits_{x\to 0}\frac{2x^5}{(x^2+1)(6x^5\cos(x^6))}\\=\lim\limits_{x\to 0}\frac{\overbrace{\sec(x^6)}^{\to 1}}{3\underbrace{(x^2+1)}_{\to 1}}=\frac 1 3$$
