Limit of the remainder of Taylor polynomial of composite functions. Problem (Spivak's Calculus, 20-9, (d)): let $f(x)=P_{n,0,f}(x)+R_{n,0,f}(x)$ and $g(x)=P_{n,0,g}(x)+R_{n,0,g}(x)$ where  $P_{n,0,f},P_{n,0,g}$ are the Taylor polynomials of degree $n$ at $0$ for $f$ and $g$, $R_{n,0,f},R_{n,0,g}$ are the corresponding remainders, and $g(0)=0$. In part of the problem I need to show that $$\lim_{x \rightarrow 0} \frac {R_{n,0,f}(g(x))} {x^n}=0$$
My attempts: Since $$\lim_{x \rightarrow 0} \frac {R_{n,0,f}(x)} {x^n}=0,$$ $P_{n,0,g}(x)$ contains only terms of degree $\geq 1$ and $R_{n,0,g}$ approaches $0$ as quickly as $x^n$, I can most likely prove this using $\epsilon - \delta$ arguments, but that seems overly complicated. I also can't use Taylor's Theorem since I don't know if $f^{(n+1)}$ exists, so I'm not really sure how I'm supposed to do this. Help would be appreciated.
Also this is in the chapter that introduces Taylor Polynomials, I've not reached anything about infinite series yet so please no solutions which involve infinite sums or expansions. Thanks in advance to the helpers :)
 A: Self-answering by request.
If $n=0$ then $R_{n,0,f}(x)=f(x)-f(0)$ and the condition is simply $$\lim_{x\rightarrow 0} f(g(x))-f(0)=0,$$ which is true since $g(0)=0$ and $f,g$ are assumed continuous (otherwise the statement is false).
For $n \geq 1$ the most straightforward solution is in terms of $O,o$ notation. Since $g$ is differentiable, $g \in O(x)$, so since $R_{n,0,f} \in o(x^n)$, $R_{n,0,f} \circ g \in o(x^n)$, which is the desired result.
Being more explicit requires $\epsilon$-$\delta$ arguments as I originally suspected. Let $\alpha=|g'(0)|+1$, then since $$\lim_{x\rightarrow 0} \frac {g(x)} {x}=\lim_{x\rightarrow 0} \frac {g(x)-g(0)} {x}=g'(0),$$ $\exists \delta_1>0$ such that if $|x|<\delta_1$, $|g(x)|<\alpha|x|$. Furthermore, since $$\lim_{x\rightarrow 0} \frac {R_{n,0,f}(x)} {x^n}=0,$$ $\forall \epsilon >0$ $\exists \delta_2>0$ such that if $|x|<\delta_2$: $$|R_{n,0,f}(x)|<\frac \epsilon {\alpha^n} |x^n|.$$ Let $\delta=\min(\delta_1,\frac {\delta_2} \alpha)$, then $\forall |x|<\delta$, $|g(x)|<\alpha|x|<\alpha \cdot \frac {\delta_2} \alpha=\delta_2$, hence:  $$|R_{n,0,f}(g(x))|<\frac \epsilon {\alpha^n} |g(x)^n|<\frac \epsilon {\alpha^n}\cdot\alpha^n\cdot|x^n|=\epsilon|x^n|.$$ Since such $\delta$ can be found $\forall \epsilon>0$, $$\lim_{x\rightarrow 0} = \frac {R_{n,0,f}(g(x))} {x^n}=0.$$
