<s>[INST] Find the limit $\lim _{x\to 0}\frac{\cos x-1+\frac{x}{2}\cdot \sin x}{\ln ^4(x+1)}$ 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
 [/INST] Write: $Q(x) =\dfrac{\cos x -1 +\dfrac{x}{2}\cdot\sin x}{\ln^4(1+x)} = \dfrac{-2\sin^2\left(\frac{x}{2}\right)+2\cdot\left(\dfrac{x}{2}\right)^2\cdot\dfrac{\sin x}{x}}{\ln^4(1+x)} = \dfrac{\dfrac{x^2}{2}}{\ln^2(1+x)}\cdot\dfrac{\left(\dfrac{\sin x}{x}-\dfrac{\sin^2\left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)^2}\right)}{\ln^2(1+x)}= f(x)g(x)$. Observe that when $x \to 0 \implies f(x) \to \dfrac{1}{2}$, and for $g(x)$, to make it simpler put $x=2y \implies g(x) = h(y) = \dfrac{\dfrac{\sin(2y)}{2y} - \dfrac{\sin^2y}{y^2}}{\ln^2(1+2y)}=\dfrac{\sin y}{y}\cdot \dfrac{\cos y - \dfrac{\sin y}{y}}{\ln^2(1+2y)}=\dfrac{\sin y}{y}\cdot \left(\dfrac{2y}{\ln(1+2y)}\right)^2\cdot \dfrac{1}{4}\cdot \dfrac{y\cos y - \sin y}{y^3}\to 1\cdot 1\cdot \dfrac{1}{4}\cdot \dfrac{-1}{3}= \dfrac{-1}{12}\implies f(x)g(x) \to \dfrac{-1}{24} \implies \displaystyle \lim_{x \to 0} Q(x) = \dfrac{-1}{24}$.
</s>