<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 only need to add to your equalities the $\log$
$$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + \textrm{h.o.t}\\
x \sin x = x^2 - \frac{x^4}{6} + \textrm{h.o.t}\\
\log(1+x) = x + \textrm{h.o.t, so} \\
\log^4(1+x) = x^4 + \textrm{h.o.t}$$
Now calculate carefully and see that
$$\frac{\cos x-1 - \frac{x}{2} \sin x}{\log^4(1+x)} =\ldots = \frac{-\frac{x^4}{24}  + \textrm{h.o.t}}{x^4 + \textrm{h.o.t}}$$
Now divide numerator and denominator by $x^4$, and get your result.  You were on the right track!
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