I need to find $\displaystyle \lim _{x\to 0}\frac{\cos x-1+\frac{x}{2}\cdot \sin x}{\ln ^4\left(x+1\right)}$. I tried using the following: \begin{align*} \ln(1+x)&\approx x,\\ \sin(x)&\approx x-\frac{x^3}{2},\\ \cos(x)&\approx 1-\frac{x^2}{2!}+\frac{x^4}{4!}, \end{align*} I managed to get $\displaystyle \lim _{x\to 0}\frac{\cos x-1+\dfrac{x}{2}\cdot \sin x}{\ln ^4\left(x+1\right)}=\lim _{x\to 0}\frac{\left(\dfrac{\left(-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}\right)}{x^{4}}+\dfrac{x^{2}-\frac{x^{4}}{3!}}{2x^{4}}\right)}{\left(\dfrac{\ln\left(1+x\right)-x}{x}+1\right)^{4}}$ but I can't see how to use that $\displaystyle \lim _{x\to 0}\frac{\ln\left(1+x\right)-x}{x}=0$ in here.
Am I missing something? I didn't learn the little-$\mathcal{O}$ notation yet.
Thanks