How to use little-o notation to solve $\lim\limits_{x\to 0} \frac{\sin{(x^2)}-\sin^2{(x)}}{\sin{(x^2)}\sin^2{(x)}}$ with Taylor polynomials? Consider the limit
$$\lim\limits_{x\to 0} \frac{\sin{(x^2)}-\sin^2{(x)}}{\sin{(x^2)}\sin^2{(x)}}\tag{1}$$
Let me preface my question by saying that I am studying the Chapter of Spivak's Calculus entitled "Approximation by Polynomial Functions", in which Taylor's Theorem is introduced. Infinite series have not been introduced yet.
I've just read the Introduction chapter of the book Asymptotic Methods in Analysis by N.G. de Bruijn to see the definitions used for the O-symbol and o-symbol.
The goal here is to solve the limit $(1)$ by using Taylor's Theorem. I asked a separate question where I show my (prolix) solution to the limit.
In that question, one person provided a much shorter answer that utilized little-o notation.
I'd like to understand a bit better how to use such notation.
Here is that solution
$$\begin{align}\lim\limits_{x\to 0} \frac{\sin{(x^2)}-\sin^2{(x)}}{\sin{(x^2)}\sin^2{(x)}}&=\lim\limits_{x\to 0} \frac{\left ( x^2-\frac{x^6}{3!}+... \right )-\left ( x-\frac{x^3}{3!}+... \right )^2}{\left ( x^2-\frac{x^6}{3!}+... \right ) \left (x-\frac{x^3}{3!}+... \right )^2}\\&\tag{2}=\lim\limits_{x\to 0} \frac{\left ( x^2+o(x^4)\right )-\left ( x^2-2x\frac{x^3}{3!}+o(x^4)\right )}{\left ( x^2+... \right ) \left (x^2+... \right )}\\&\tag{3}=\lim\limits_{x\to 0}\frac{\frac{2x^4}{3!}}{x^4}\\
&=\frac13.\end{align}$$

*

*What happened to little-o in the numerator between $(2)$ and $(3)$? At first sight it would seem we have $o(x^4)-o(x^4)$, but as far as I can tell the entity $o(x^4)$ can't be used in a subtraction like this (ie we can't just cancel them, can we?).

*Why wasn't little-o used in the denominator?

*More generally, what exactly (if possible in painstaking detail) do we do to get from $(2)$ to $(3)$?

 A: *

*$o(x^4)$ means "any function $f(x)$ such that $\frac{f(x)}{x^4}\to0$" (here: when $x\to0$). So $o(x^4)-o(x^4)$ means "any difference of any two such functions" (which may be $\ne$), and such a difference is again some $o(x^4)$.

*My lazy writing was only to imitate the style of Spivak's solution you reported in your previous question. More rigorously, this becomes
$$\begin{align}\lim\limits_{x\to 0} \frac{\sin{(x^2)}-\sin^2{(x)}}{\sin{(x^2)}\sin^2{(x)}}&=\lim\limits_{x\to 0} \frac{\left ( x^2+o(x^4)\right )-\left ( x-\frac{x^3}{3!}+o(x^3)\right )^2}{\left ( x^2+o(x^2)\right ) \left (x+o(x)\right )^2}\\&\tag{2}=\lim\limits_{x\to 0} \frac{\left ( x^2+o(x^4)\right )-\left ( x^2-2x\frac{x^3}{3!}+o(x^4)\right )}{\left ( x^2+o(x^2)\right ) \left (x^2+o(x^2)\right )}\\&\tag{3}=\lim\limits_{x\to 0}\frac{\frac{2x^4}{3!}}{x^4}\\
&=\frac13.\end{align}$$

*Using the notation $o(1)$ for "any function $f$ which tends to $0$", we have $o(x^4)=x^4o(1)$, hence$$\begin{align}\tag{2}\lim\limits_{x\to 0} \frac{\left ( x^2+o(x^4)\right )-\left ( x^2-2x\frac{x^3}{3!}+o(x^4)\right )}{\left ( x^2+o(x^2)\right ) \left (x^2+o(x^2)\right )}&=\lim\limits_{x\to 0} \frac{\frac{2x^4}{3!}+o(x^4)}{x^4+o(x^4)}\\&=\lim\limits_{x\to 0} \frac{\frac{2x^4}{3!}(1+o(1))}{x^4(1+o(1))}\\&=\lim\limits_{x\to 0} \frac{\frac{2x^4}{3!}}{x^4}\lim\limits_{x\to 0} \frac{1+o(1)}{1+o(1)}\\&\tag{3}=\lim\limits_{x\to 0}\frac{\frac{2x^4}{3!}}{x^4}.\end{align}$$
But this can be written more shortly, using the notion of asymptotic equivalence.

A: Here is the general outline, which you can adapt to the given Case.
When we have $L=\lim_{x \rightarrow 0}{\frac{A(x)}{B(x)}}$ [[ EQ 1 ]]
& we eventually get :
$A(x)=a_{n}x^{n}+a_{n+1}x^{n+1}+a_{n+2}x^{n+2}+a_{n+3}x^{n+3}+\cdots$
$B(x)=b_{n}x^{n}+b_{n+1}x^{n+1}+b_{n+2}x^{n+2}+b_{n+3}x^{n+3}+\cdots$
where both are starting at some Power $n$ ,
then we can take out $x^{n}$ to get :
$A(x)=x^{n}(a_{n}x^{0}+a_{n+1}x^{1}+a_{n+2}x^{2}+a_{n+3}x^{3}+\cdots)$
$B(x)=x^{n}(b_{n}x^{0}+b_{n+1}x^{1}+b_{n+2}x^{2}+b_{n+3}x^{3}+\cdots)$
Hence $L=\lim_{x \rightarrow 0}{\frac{x^{n}(a_{n}x^{0}+a_{n+1}x^{1}+a_{n+2}x^{2}+a_{n+3}x^{3}+\cdots)}{x^{n}(b_{n}x^{0}+b_{n+1}x^{1}+b_{n+2}x^{2}+b_{n+3}x^{3}+\cdots)}}$ [[ EQ 2 ]]
Naturally, we can cancel $x^{n}$ in Numerator & Denominator & substitute $x=0$ to get the limit $L=\frac{a_{n}+0+0+0+\cdots}{b_{n}+0+0+0+\cdots}$
The higher Power terms will be $0$ in the limit, which we can simply write $o(x^{n})$ in EQ 1 or $o(x)$ in EQ 2.
$A(x)=a_{n}x^{n}+o(x^{n})$
$B(x)=b_{n}x^{n}+o(x^{n})$
In case the original limit is more generally like this :
$L=\lim_{x \rightarrow 0}{\frac{A(x)-U(x)}{B(x)V(x)}}$ [[ EQ 3 ]]
Where :
$A(x)=a_{2n}x^{2n}+o(x^{2n})$
$B(x)=b_{n}x^{n}+o(x^{n})$
$U(x)=u_{2n}x^{2n}+o(x^{2n})$
$V(x)=v_{n}x^{n}+o(x^{n})$
then we will get :
$L=\lim_{x \rightarrow 0}{\frac{(a_{2n}x^{2n}+o(x^{2n}))-(u_{2n}x^{2n}+o(x^{2n}))}{(b_{n}x^{n}+o(x^{n}))(v_{n}x^{n}+o(x^{n}))}}$ [[ EQ 4 ]]
We can not cancel the $o(\cdot)$ terms but we can multiply, to get higher terms.
What this means :
$o(x^{n})+o(x^{n})=o(x^{n})$ (basically the same set)
$o(x^{n})-o(x^{n})=o(x^{n})$ (these will not cancel !)
$o(x^{n}) \times o(x^{m})=o(x^{n+m})$ (add the Power of each term)
$o(x^{n}) / o(x^{m})=o(x^{n-m})$ (subtract the Power $m$ which is less than the Power $n$)
Using that, we get :
$L=\lim_{x \rightarrow 0}{\frac{a_{2n}x^{2n}-u_{2n}x^{2n}+o(x^{2n})}{b_{n}v_{n}x^{2n}+o(x^{2n})}}$ [[ EQ 5 ]]
$L=\lim_{x \rightarrow 0}{\frac{x^{2n}(a_{2n}-u_{2n}+0)}{x^{2n}(b_{n}v_{n}+0)}}$ [[ EQ 6 ]]
where we cancel $x^{2n}$ in Numerator & Denominator & substitute $x=0$ to get the limit with the terms $a_{2n}, u_{2n}, b_{n}, v_{n}$ which will not vanish in the limit.
