The first part of the exercise requires stating the order of: $$h(x)=\lim_{x\to0} (1+4x)^{1/4} + \ln(1-\sin(x)) - 1$$
I approximated $(1+4x)^{1/4}$ as $1/4\times4x$, that is $x$, and $\ln(1-\sin(x))$ can be approximated as $x$.
(That is, $-\sin(x)\sim -x$, and $\ln(1-x)\sim -x$ for $x$ sufficiently small.)
I'm conflicted because according to the solution my teacher put online, it should be an infinitesimal of the second order. I noticed she used little-o notation but I couldn't make heads or tails as to why there was an $x^2$ and $o(x^2)$ in there.
The other part of the exercise requires calculating $\lim_{x\to0}\frac{1-\cos(3x)}{h(x)}$ and I can see that equaling a constant only if $h(x)$ is an infinitesimal of the second order ($1-\cos(x)\sim \frac{1}{2}x^2)$.
We have that
$$(1+4x)^{1/4}=1+x-\frac32x^2+o(x^2)$$
$$\ln(1-\sin(x))=\ln(1-x+o(x^2))=-x-\frac12x^2+o(x^2)$$
therefore
$$(1+4x)^{1/4}+\ln(1-\sin(x))-1=-2x^2+o(x^2)$$
and then for the second part
$$\frac{1-\cos(3x)}{h(x)} =\frac{\frac92x^2+o(x^2)}{-2x^2+o(x^2)}=\frac{\frac92+o(1)}{-2+o(1)}\to -\frac94$$