I've been looking at the solution to this question for some time and I can't workout why the coefficients of $x^4$, $x^6$ and $x^8$ are $0$. I'm also wondering where the $x^4/2$ came from in the 2nd line. I know that in the solution the Taylor polynomials of the terms are used but that's about all I know.
Here is the solution: \begin{align*} \lim_{x\to0}\frac{x^2-\cos(x)\sin(x^2)}{\ln(1+2x^4)}&= \lim_{x\to0}\frac{x^2-(1-\frac{x^2}{2}-\mathcal O (x^4))(x^2+\mathcal O (x^6))}{2x^4+\mathcal O(x^8)}\\ &=\lim_{x\to 0}\frac{\frac{x^4}{2}+\mathcal O(x^6)}{2x^4+\mathcal O(x^8)}\\ &=\lim_{x\to 0}\frac{\frac{1}{2}+\mathcal O (x^2)}{2+\mathcal O (x^4)}\\ &=\frac{1}{4}. \end{align*}
Thanks