I had an exercise requiring finding the infinitesimal order of $h(x)=(1+4x)^\frac{1} {4}+ln(1−sin(x))−1$ for $x$ approaching $0$, and calculating $\lim_{x\to0}\frac{1-\cos(3x)}{h(x)}$.
The correct answer as stated by the professor and people on math.stackexchange would be:
$$(1+4x)^{1/4}=1+x-\frac32x^2+o(x^2)$$
$$\ln(1-\sin(x))=\ln(1-x+o(x^2))=-x-\frac12x^2+o(x^2)$$
Resulting in:
$$(1+4x)^{1/4}+\ln(1-\sin(x))-1=-2x^2+o(x^2)$$
and $$\frac{1-\cos(3x)}{h(x)} =\frac{\frac92x^2+o(x^2)}{-2x^2+o(x^2)}=\frac{\frac92+o(1)}{-2+o(1)}\to -\frac94$$ for the second part.
What confuses me is why specifically the quadratic approximation and not say the cubic one?
I know $\lim_{x\to0}\frac{1-\cos(3x)}{h(x)}$ can be a constant only if $h(x)$ is of the second order (as $1-\cos(3x)\sim \frac{9}{2}x^2)$.
If we did the cubic approximation, the $x^3$'s wouldn't cancel out. I feel lost because I can find no reliable sources online and my textbooks are very sparse when it comes to explanations. Hope it is clear for you guys.
Let's suppose we do not stop at the $x^2$ term. Instead we include the exact $x^3$ term, leaving only $o(x^3)$ at the end. And now we want the limit
$$ \lim_{x\to 0} \frac{ax^2 + cx^3 + o(x^3)}{bx^2 + dx^3 + o(x^3)}, $$
where $a,b,c,d$ are whatever Taylor coefficients we obtained. (In your problem $a = \frac92$ and $b = -2.$)
It is true that we cannot cancel $x^3.$ But we can still cancel $x^2,$ and when we do that, we have to cancel $x^2$ from the entire expression on top and the entire expression on the bottom, from every term in that expression.
That is,
\begin{align} \lim_{x\to 0} \frac{ax^2 + cx^3 + o(x^3))}{bx^2 + dx^3 + o(x^3))} &= \lim_{x\to 0} \frac{x^2(a + cx + o(x))}{x^2(b + dx + o(x))}\\ &= \lim_{x\to 0} \frac{a + cx + o(x)}{b + dx + o(x)}. \end{align}
Now we know that if $\lim_{x\to 0} f(x) = a$ and $\lim_{x\to 0} g(x) = b \neq 0,$ that
$$ \lim_{x\to 0} \frac{f(x)}{g(x)} = \frac ab. $$
And that is exactly what happens here, because $\lim_{x\to 0} a + cx + o(x) = a$ and $\lim_{x\to 0} b + dx + o(x) = b.$
Notice that it doesn't matter what we find for $c$ and $d.$ Their values have no effect on the limit value. So any effort spent calculating them is useless for the purpose of this particular limit. That is presumably why your professor simply lumped them into the $o(x^2)$ rather than calculating exact terms to any higher order.