Exercise 3 in Section 7.11 of Apostol's Calculus (Vol. 1) little o-notation The problem is stated as:

Find the polynomial $P(x)$ of minimal degree such that
  $$
\sin(x-x^2)=P(x)+o(x^6)\quad \textrm{as }x\to 0.
$$

and the answer in the book is

$$
P(x)=x-x^2-\frac{x^3}{6}+\frac{x^4}{2}-\frac{59x^5}{120}+\frac{x^6}{8}
$$

I am wondering why the answer cannot be 
$P(x) = x-x^2$
Since we can use the fact that $\sin(x-x^2) = x-x^2 + o((x-x^2)^2)$ as $x \to 0$
And we know that $o((x-x^2)^2) = o(x^5) = o(x^6)$ as $x$ goes to zero.
Then why is the answer from the back of the book not equal to $x-x^2$??
 A: What they are asking you is to find a polynomial $p(x)$ such that
$$
\lim_{x\rightarrow0}\frac{|\sin(x-x^2)-p(x)|}{x^6}=0
$$
No to your statements: 


*

*$o((x-x^2)^2) =o(x^5)$ is in general false. For example $h(x)=x(x-x^2)^2=o((x-x^2)^2$ as $x\rightarrow0$ but $x^{-5}h(x)\nrightarrow0$ as $x\rightarrow0$. 

*$o(x^5) = o(x^6)$ is false in general: $g(x):=x^6=o(x^5)$ but $g(x) \neq o(x^6)$.


To help you solve the problem, use Taylor expansion of $\sin z= z-\frac{z^3}{6}+\frac{z^5}{5!}-\frac{z^7}{7!}+...$. Notice that when $z=x-x^2$, $z^7$ is a polynomial on $x$ of degree 14 of the $z^7=x^7-7x^8+...-x^{14}$ This term is of course $o(x^6)$ since all powers involved are larger than $6$. So truncating the sine series up to the $7-th$ power gives the right decay of residual. The rest is to check that the 
$$
(x-x^2)-\frac{(x-x^2)^3}{6}+\frac{(x-x^2)^5}{5!}
$$
is the polynomial in the statement of your problem. A little but tedious algebra.
A: A sketch of the correct method:
Since $x-x^2$ has order $1$, you have to expand $\sin u$ up to degree $6$, and substitute $x-x^2$ in $u, u^2,\dots u^6$, truncating the expansions of the different $(x-x^2)^k$ at degree $6$, which leads to a recursive computation:
\begin{align}
(x-x^2)^2&=x^2-2x^3+x^4& \quad(x-x^2)^3&=x^3-3x^4+3x^5-x^6 \\
(x-x^2)^4&=x^4-4x^5+6x^6-4x^7+x^8&(x-x^2)^5&=(x-x^2)^4(x-x^2)\\
&=x^4-4x^5+6x^6+o(x^6)&&=\bigl((x^4-4x^5+6x^6+o(x^6)\bigr)(x-x^2)\\
(x-x^2)^6&=(x-x^2)^5(x-x^2)&&=\begin{aligned}[t]&x^5-4x^6+6x^7+o(x^7)\\&\phantom{x5}-x^6+4x^7-6x^8+o(x^8)
&\end{aligned}\\&=\bigl(x^5-5x^6+o(x^6)\bigr)(x-x^2)&&=x^5-5x^6+o(x^6)\\
&=\dots=x^6+o(x^6)
\end{align}
Can you end the computations?
A: Note that one has $o(x^6)=o(x^5)$ (as $x\to 0$), but not $o(x^5)=o(x^6)$. 
Also, $o((x-x^2)^2)=o(x^5)$ is incorrect: one has $\lim_{x\to 0}\frac{x^5}{(x-x^2)^2}=0$ but $\lim_{x\to 0}\frac{x^5}{x^5}=1$. In general, $o(f(x))=o(g(x))$ does NOT imply that $o(g(x))=o(f(x))$ as discussed in an answer to another question of yours.

Why the answer cannot be $x-x^2$?

It is an instructive exercise to show (using the correct answer in Apostol) that the following is FALSE:
$$
\lim_{x\to 0}\frac{\sin (x-x^2)-(x-x^2)}{x^6}= 0\tag{1}
$$

[Added.]
To see why (1) is false, you can use the solution of $P(x)$ in the book
so that
$$\begin{align}
f(x):=\frac{\sin (x-x^2)-(x-x^2)}{x^6}
&=\frac{a_3x^3+a_4x^4+a_5x^5+a_6x^6+o(x^6)}{x^6}\\
&=\frac{1}{x^3}
\left(
a_3+a_4x+a_5x^2+a_6x^3
\right)+\frac{o(x^6)}{x^6}\\
&=\frac{1}{x^3}\cdot g(x)+h(x)
\end{align}
$$
where $g(x):=a_3+a_4x+a_5x^2+a_6x^3$.
Now, note that
$$
\lim_{x\to 0}g(x)=a_3\neq 0,\quad \lim_{x\to 0}\frac{o(x^6)}{x^6}=0.
$$
So
$
\lim_{x\to 0+}f(x)=\infty
$
and (1) is thus impossible.
