How to evaluate $\lim _{x\to \:0}\left(\frac{\cos \left(x\right)-1+\frac{x}{2}\sin \left(x\right)}{\ln ^4\left(1+x\right)}\right)$

Question

I try using the asymptotic equivalence, but it bring to the cancellation of the numerator $$\lim _{x\to \:0}\left(\frac{\cos \left(x\right)-1+\frac{x}{2}\sin \left(x\right)}{\ln ^4\left(1+x\right)}\right)\approx \lim _{x\to 0}\left(\frac{1-\frac{x^2}{2}-1+\frac{x^2}{2}}{x^4}\right)$$ How can i solve it?

Answer 1

$$\lim _{x\to \:0}\left(\frac{\cos \left(x\right)-1+\frac{x}{2}\sin \left(x\right)}{\ln ^4\left(1+x\right)}\right)$$ You have to use the taylor expansion, because the asymptotic equivalence does not offer a fine local approximation enough to solve the following limit.
So:
$\cos(x) = 1-\frac{1}{2!}x^2+\frac{1}{4!}x^4 + o(x^4)$
$\sin(x) = \frac{1}{1!}x-\frac{1}{3!}x^3 + o(x^3)$
$\ln(1+x) = x+o(x)$
$$\lim _{x\to \:0}\left(\frac{\cos \left(x\right)-1+\frac{x}{2}\sin \left(x\right)}{\ln ^4\left(1+x\right)}\right)= \lim _{x\to 0}\left(\frac{1-\frac{1}{2!}x^2+\frac{1}{4!}x^4\:+\:o\left(x^4\right)-1+\frac{x}{2}\left(\frac{1}{1!}x-\frac{1}{3!}x^3\:+\:o\left(x^3\right)\right)}{\left(x+o\left(x\right)\right)^4}\right) = \lim _{x\to 0}\left(\frac{-\frac{x^4}{24}+\:o\left(x^4\right)}{\left(x+o\left(x\right)\right)^4}\right) \rightarrow\color{red}{-1/24}$$

Answer 2

Hint. Your expansions at the numerator need more terms. Try with $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{4!}+o(x^4)\quad \mbox{and}\quad \sin(x)=x-\frac{x^3}{3!}+o(x^3).$$ Then as $x\to 0$, $$\frac{\cos \left(x\right)-1+\frac{x}{2}\sin \left(x\right)}{\ln ^4\left(1+x\right)}= \frac{?}{x^4+o(x^4)}\to ?$$