Computing limits using Taylor expansions and $o$ notation on both sides of a fraction Let's define $o(g(x))$ as usually:
$$
\forall x \ne a.g(x) \ne 0 \\
f(x) = o(g(x)) \space \text{when} \space x \to a \implies \lim_{x \to a} \frac{f(x)}{g(x)}=0
$$
In theorem $7.8$, Tom Apostol in his Calculus Vol. $1$ gave and proved the following basic rules for algebra of o-symbols

*

*$o(g(x)) \pm o(g(x)) = o(g(x)) $

*$o(cg(x)) = o(g(x)) $, if $c \ne 0$

*$f(x) \cdot o(g(x)) = o(f(x)g(x)) $

*$o(o(g(x))) = o(g(x)) $

*$\frac{1}{1 + g(x)} = 1 - g(x) + o(g(x))$
I will use a concrete example to demonstrate the confusion I have, but the question is probably more generally applicable. By using Taylor expansions, compute $\lim_{x \to 0} \frac{1 - \cos x^2}{x^2 \sin x^2}$.
\begin{equation}
\cos x^2 = 1 - \frac{x^4}{2} + o(x^5) \\
\sin x^2 = x^2 + o(x^4)
\end{equation}
$$
\lim_{x \to 0} \frac{1 - \cos x^2}{x^2 \sin x^2} = \lim_{x \to 0} \frac{1 - (1 - \frac{x^4}{2} + o(x^5))}{x^2 (x^2 + o(x^4))} = \lim_{x \to 0} \frac{\frac{x^4}{2} + o(x^5)}{x^4 + o(x^6)}
$$
What would be an easy way to solve that, without flattening the fraction by the application of case $5$ of the theorem $7.8$ above, while using the provided definition for $o$ notation (and without using a more advanced methods, like L'Hopital's rule)?
I saw somewhere $\lim_{x \to 0} \frac{\frac{1}{2} + \frac{o(x^5)}{x^4}}{1 + \frac{o(x^6)}{x^4}} = \frac{1}{2}$, without explanation, suggesting it should be a trivial matter, but I'm not sure why that would be the case.

Appendix in how I solved that, relying only on the definition and the $T7.8$.
One way to proceed could be to use case $3$ from Theorem $7.8$, in reverse (i.e. $o(x^6) = x^4 o(x^2)$)
$$
\lim_{x \to 0} \frac{\frac{x^4}{2} + o(x^5)}{x^4 + o(x^6)} = \lim_{x \to 0} \frac{\frac{1}{2} + o(x)}{1 + o(x^2)}
$$
Then I could apply case $5$ from the above theorem and applying case $4$ of the theorem (i.e. $o(o(x^2)) = o(x^2)$), to get $\frac{1}{1+o(x^2)} = 1 - o(x^2)$.
After applying case $2$ with $c = -1$, we get:
$$
\lim_{x \to 0} \frac{\frac{1}{2} + o(x)}{1 + o(x^2)} = \lim_{x \to 0} (\frac{1}{2} + o(x)) (1 + o(x^2)) = \frac{1}{2}
$$
 A: Observe that
$$
\frac{o(x^5)}{x^4} = \frac{o(x^5)}{x^5}x \to 0 \quad \text{as } x \to 0$$
since, by definition of $o$, we have $o(x^5)/x^5 \to 0$ as $x \to 0$.
Similarly,
$$
\frac{o(x^6)}{x^4} = \frac{o(x^6)}{x^6}x^2 \to 0 \quad \text{as } x \to 0.
$$
This explains why
$$
\lim\limits_{x \to 0} \frac{\frac{1}{2}+\frac{o(x^5)}{x^4}}{1+\frac{o(x^6)}{x^4}} = 0.
$$
Moreover, the above computations show also that $o(x^5) = o(x^4)$ and $o(x^6)= o(x^4)$ as $x\to 0$.
So you could also write
$$
\begin{align}
\lim_{x \to 0} \frac{\frac{x^4}{2} + o(x^5)}{x^4 + o(x^6)} &= \lim_{x \to 0} \frac{\frac{x^4}{2} + o(x^4)}{x^4 + o(x^4)} \\
& = \lim_{x \to 0} \frac{\frac{1}{2}+o(1)}{1+o(1)} = \frac{1}{2}
\end{align}
$$
A: A half is the correct answer. For instance, something that is smaller than $ x^5 $ or $ x^6 $ is neglected in comparison to something that grows (or decreases) at $x^4$.
Remember that $o(x)$ is just a notation for something else, it may be a large polynomial approximation, so as it happens to be in this case.
