Does $o(x)$ show the order of Taylor expansion? If we have :
$$
\sin x= x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !} +o\left(x^{5}\right)
$$
and $$
\sin x=
x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}
+o\left(x^{6}\right).
$$
Is the order of the first expression of $\sin(x)$ 5, and the order of the second one $6$?
If so, why then can we replace $o(x^6)$ with $o(x^5)$?
I know it's because as $x$ tends to $0$, $o(x^6)$ is a subset of $o(x^5)$ but won't the order change if we replace $o(x^6)$ with $o(x^5)$?
 A: The little-o means the following:

If $f\in o(g(x))$, as $x\to a$ (here the value $a$ is $0$; with Taylor series in general $a$ will be the centrepoint of the expansion), then: $$\lim_{x\to a}\frac{f(x)}{g(x)}=0$$

Whereas Big-O is the weaker requirement of:
$$\lim_{x\to a}\frac{f(x)}{g(x)}\lt\infty$$
In Taylor expansions, you'll see either of the two. For example, the remainder terms of sine after a $5$th order expansion are:
$$-\frac{x^7}{7!}+\frac{x^9}{9!}-\cdots$$
Which is both $o(x^5)$ and $o(x^6)$, as $x\to0$, because if we divide all terms by $x^5$ or $x^6$, we'll still have powers of $x$, which go to zero as $x\to0$. However, you might also see the remainder written as $O(x^7)$. The Big-O means the ratio is finite; if we divide that remainder by $x^7$, you'll notice that we will have:
$$-\frac{1}{7!}+\frac{x^2}{9!}-\cdots$$
Which goes to $-\frac{1}{7!}\lt\infty,\,x\to0$.
Usually, the little-o will correspond to the order of the expansion, and the Big-O to the order of the remainder, but that isn't what these symbols actually are defined to mean, so bear in mind the definitions. In this case, the $5$th and $6$th order expansions of sine are the same, which is maybe a more intuitive reason why the two little-o's were the same.
