<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] You can use
$$ \lim_{x\to0}\frac{\ln(1+x)}{x}=1 $$
to handle the limit. In fact
\begin{eqnarray}
&&\lim _{x\to 0}\frac{\cos x-1+\frac{x}{2}\cdot \sin x}{\ln ^4\left(x+1\right)}\\
&=&\lim _{x\to 0}\frac{\cos x-1+\frac{x}{2}\cdot \sin x}{x^4}\bigg[\frac{x}{\ln \left(x+1\right)}\bigg]^4\\
&=&\lim _{x\to 0}\frac{-\sin x+\frac{1}{2} \sin x+\frac{x}{2}\cos x}{4x^3}\\
&=&\lim _{x\to 0}\frac{-\frac{1}{2} \cos x+\frac{1}{2}\cos x-\frac{x}{2}\sin x}{12x^2}\\
&=&\lim _{x\to 0}\frac{-\sin x}{24x}\\
&=&-\frac{1}{24}.
\end{eqnarray}
</s>