Spivak, Ch. 20, Problem 9c: Is there a Typo in this question item or not? The following problem is from Chapter 20 of Spivak's Calculus. I've asked a separate question regarding the comment at the end of item $(c)$. In the accepted answer to that question, the author of the response claims that item $(c)$ itself has a typographical error. I had already come up with a proof for that item, however, and my question now is if such a proof is incorrect. In other words, was the author of the aforementioned answer correct in his claim.



*(a) Problem $7(i)$ amounts to the equation

$$P_{n,a,f+g}=P_{n,a,f}+P_{n,a,g}$$
Give a more direct proof by writing
$$f(x)=P_{n,a,f}(x)+R_{n,a,f}(x)\tag{1}$$
$$g(x)=P_{n,a,g}(x)+R_{n,a,g}(x)\tag{2}$$
and using the obvious fact about $R_{n,a,f}+R_{n,a,g}$.
(b) Similarly, Problem $7(ii)$ could be used to show that
$$P_{n,a,fg}=[P_{n,a,f}\cdot P_{n,a,g}]_n$$
where $[P]_n$ denotes the truncation of $P$ to degree $n$, the sum of
all terms of $P$ of degree $\leq n$ [with $P$ written as a polynomial
in $x-a$]. Again, give a more direct proof, using the obvious facts
about products involving terms of the form $R_n$.
(c) Prove that if $p$ and $q$ are polynomials in $x-a$ and
$\lim\limits_{x\to 0} \frac{R(x)}{(x-a)^n}=0$ then
$$p(q(x)+R(x))=p(q(x))+\bar{R}(x)$$
where $$\lim\limits_{x\to 0} \frac{\bar{R}(x)}{(x-a)^n}=0$$

The claimed typo seems to be that $x\to a$ in the limits instead of $x\to 0$.
My attempt at a proof
$$p(x)=\sum\limits_{i=1}^n a_i(x-a)^i$$
$$p(q(x)+R(x))=\sum\limits_{i=0}^n a_i(q(x)+R(x)-a)^i=\sum\limits_{i=0}^n a_i \sum\limits_{j=0}^i \binom{i}{j} (q(x)-a)^{i-j} R(x)^j$$
$$=\sum\limits_{i=0}^n a_i(q(x)-a)^i + \sum\limits_{i=0}^n a_i \sum\limits_{j=1}^i \binom{i}{j}(q(x)-a)^{i-j}R(x)^j$$
$$=p(q(x))+\bar{R(x)}$$
where $\bar{R}(x)=\sum\limits_{i=0}^n a_i \sum\limits_{j=1}^i \binom{i}{j}(q(x)-a)^{i-j}R(x)^j$
Note that $(q(x)-a)^{i-j}$ is polynomial, hence continuous everywhere and $\lim\limits_{x\to 0} (q(x)-a)^{i-j}$ exists.
Thus, $\lim\limits_{x\to 0} \frac{\bar{R}(x)}{(x-a)^n}=\sum\limits_{i=0}^n a_i \sum\limits_{j=1}^i \binom{i}{j} \cdot \lim\limits_{x\to 0} (q(x)-a)^{i-j} \cdot \frac{R(x)^j}{(x-a)^n}=0$
Is this proof correct?
 A: Your argument is fine, there is a typo in the book, and you are maybe missing the point.
Note that for $a\ne0$ the condition $$\tag1\lim\limits_{x\to 0} \frac{R(x)}{(x-a)^n}=0$$ is equivalent with $$\tag2R(0)=0.$$ It's a bit cumbersome to write $(2)$ as $(1)$.
The point of Taylor polynomials around $a$ is that you want $$\tag3\lim\limits_{x\to a} \frac{R(x)}{(x-a)^n}=0$$ to hold. That is, you want the difference $f(x)-p(x)$ to be $o((x-a)^n)$. This means precisely that
$$\tag4\lim\limits_{x\to a} \frac{f(x)-p(x)}{(x-a)^n}= \lim\limits_{x\to a} \frac{R(x)}{(x-a)^n}=0.$$
A: If you replace every $x\to0$ by $x\to a$ as Martin Argerami explained, and expand $(q(x)+R(x)-a)^i$ only for $i\ge1$, your proof is correct except that $p(a)$ is neither supposed to be $0$ nor $p$ to be of degree $\le n.$ (Note that Spivak's assumption that $q$ is a polynomial is also useless: if $i>j\ge1$, for $\lim\limits_{x\to a} (q(x)-a)^{i-j} \cdot \frac{R(x)^j}{(x-a)^n}$ to be $0$, boundedness of $q$ around $a$ is sufficient.)
The proof can be shortened by writing $p(t)=\sum_{k=0}^db_kt^k$ and using little o notation and its properties:
As $x\to a$, if $R(x)=o((x-a)^n)$ then $$\begin{align}p(q(x)+R(x))&=\sum_{k=0}^db_k\left(q(x)+o((x-a)^n)\right)^k\\&=b_0+\sum_{k=1}^db_k\left((q(x))^k+o((x-a)^n)\right)\\&=\left(\sum_{k=0}^db_k(q(x))^k\right)+\left(\sum_{k=1}^db_ko((x-a)^n)\right)\\&=p(q(x))+o((x-a)^n).\end{align}$$
Edit: to justify the replacement of $\left(q(x)+o((x-a)^n)\right)^k,$ for $k\ge1,$ by $(q(x))^k+o((x-a)^n),$ I implicitely used the binomial theorem like you and the (local) boundedness of $q$, and the replacement of $o((x-a)^{nk})$ by $o((x-a)^n)$, which is legit since $nk\ge n$ (this is what I meant by "little o notation and its properties").
