Spivak Problem - Limit of the remainder of a Taylor polynomial of composed functions Problem (Spivak's Calculus, 20-9, (d)): The problem says "Let $a=0$ and $b=g(0)=0$, prove that $$P_{n,a,f\circ g}=\left [ P_{n,b,f}\circ P_{n,a,g} \right ]_{n}.$$ In this context, $\left [ P\right ]_{n}$ indicates the truncation of $P$ until the degree $n$; that is, the sum of all the terms of $P$ of degree $\leq n$, with $P$ written as a polynomial on $(x-a)$. "
In one part of the solution, I need to prove that $$\lim_{x\rightarrow 0}\frac{R_{n,0,f}(g(x))}{x^{n}}=0.$$ I take a look at the solution book and the first problem that I found is that the solution assumes that the $f^{(n+1)}$ and $g^{(n+1)}$ derivatives exist; a condition which is not in the problem statement, and then applies Taylor's Remainder theorem. The second problem is that the solution says "Note that $R_{n,0,f}$ is a polynomial that have only terms of degree $> n$ and $g(x)$ is a polynomial whose constant degree is $b=g(0)=0$, then all the terms of $R_{n,0,f}(g(x))$ are of degree $>n$ and it follows that $$\lim_{x\rightarrow 0}\frac{R_{n,0,f}(g(x))}{x^{n}}=0."$$
In this part, I am completely lost because, why the solution says that $R_{n,0,f}$ and $g(x)$ are polynomials? When clearly $g(x)$ is not necessary a polynomial, by the conditions, and also $R_{n,0,f}$ is not a polynomial because, even though, it can be written as $$R_{n,0,f}=\frac{f^{(n+1)}(t)}{(n+1)!}x^{n}$$
The term $t$ depends of $x$ ; that is, $f^{(n+1)}(t)$ is not a constant when $x$ goes to zero.
The only way that I can see how to prove this is by assuming that the $f^{(n+1)}$ and $g^{(n+1)}$ derivatives are bounded, then I can bound $R_{n,0,f}$ and $g(x)$ by polynomials; and then I can prove that  $$\lim_{x\rightarrow 0}\frac{R_{n,0,f}(g(x))}{x^{n}}=0."$$ by the Sandwich theorem.
I will really appreciate any help.
 A: I ran into the same issue and got through it in the following way. I feel I might be overlooking an easier method, but I think it works.
Let $R=R_{n,0,f}$. We have that $R(0)=0$, $g(0)=0$, and $\lim_{x\to 0} \frac{R(x)}{x^n}=0$ because it is the $n$-th Taylor remainder. I will also need that $g$ is differentiable at $0$, and this assumes $n \geq 1$ but the $n=0$ case can be proved separately by continuity.
I have two cases, $g'(0)\neq 0$, $g'(0)=0$.
Case 1: If $g'(0)\neq 0$, then $g\neq 0$ in some interval around $0$. By a limit change of variable, we get that $\lim_{x\to 0} \frac{R(g(x))}{g(x)^n} = 0$. Since $g$ is differentiable at $0$, we have $\lim_{x\to 0} \frac{g(x)^n}{x^n} = (g'(x))^n$. By product of limits, we get that $\lim_{x\to 0} \frac{R(g(x))}{x^n} = 0$.
Case 2: Suppose $g'(0)=0$. Let $\epsilon >0$. Then:

*

*Pick $\delta_1 > 0 $ such that $0<|x|<\delta_1 \implies |\frac{R(x)}{x^n}| < \epsilon$.

*Pick $\delta_2 >0$ such that $0<|x|<\delta_2 \implies |\frac{g(x)}{x}|< \min(1,\delta_1)$.

*Let $0<|x|<\min\{\delta_1,\delta_2\}$.

*

*If $g(x)\neq 0$, then $|\frac{R(g(x))}{g(x)^n}|<\epsilon$. Then $|\frac{R(g(x))}{g(x)^n}| |\frac{g(x)^n}{x^n}| < \epsilon$.

*If $g(x)=0$, then $|\frac{R(g(x))}{x^n}| = |\frac{R(0)}{x^n}| = 0 < \epsilon$.



*It follows that $\lim_{x\to 0} \frac {R(g(x))}{x^n} = 0$.

