understanding a proof of a function lying in the closure of a polynomial space 
Show that $f(x) = |x|$ lies in the closure of $\mathcal{P}[-1, 1]$ in $\mathcal{C}[-1,1]$ (using the supremum metric), where for $a,b\in\mathbb{R}, \mathcal{P}[a,b]$ is the set of polynomials on $[a,b]$ and $\mathcal{C}[a,b]$ is the set of continuous functions on $[a,b].$

Below is a proof of this proposition.


I understand most of it, except the claim that $|\frac{g^{(n+1)}(t)}{(n+1)!}(x-1)^{n+1}| \leq \dfrac{1\cdot 3 \cdots (2n-1)\cdot a^{2n-1}}{2^{n+1}{(n+1)!}}.$ Can someone explain why this inequality holds?

 A: The Lagrange form of the remainder of the Taylor series is
$$R_n(x) =  \frac{g^{n+1}(t(n,x))}{(n+1)!}(x-1)^{n+1}  = \frac{r_n \cdot (x -1)^{n+1}}{(t(n,x)+a^2)^{(2n+1)/2}}$$
where $r_n = \dfrac{1\cdot3 \cdot \ldots \cdot (2n-1)}{2^{n+1}(n+1)!}$ and $t(n,x) \in (x,1)$. We want to show that
$$d_n := \sup_{x \in [0,1]} \lvert R_n(x) \rvert \to 0$$
as $n \to \infty$. But certainly
$$\sup_{x \in [0,1]} \lvert R_n(x) \rvert \ge  \lvert R_n(0) \rvert = \frac{r_n}{(t(n,0)+a^2)^{(2n+1)/2}} .$$
As long as we do not know anything about $t(n,0) \in (0,1)$, we cannot exclude that there exists $b < 1$ such that $t(n,0) + a^2 \le b^2$ for all $n$. In that case
$$d_n \ge \frac{r_n}{b^{2n+1}} .$$
But we have
$$r_n  \ge \dfrac{2 \cdot 4 \cdot \ldots \cdot (2n-2)}{2^{n+1}(n+1)!} = \dfrac{2^{n-1} (n-1)!}{2^{n+1}(n+1)!} = \dfrac{1}{4n(n+1)}$$
and therefore
$$d_n \ge \frac{1}{4n(n+1)b^{2n+1}} .$$
The RHS diverges to $+\infty$.
This does not necessarily mean that the claim $d_n \le \dfrac{1}{2(n+1)}$ is false, but a proof would require to know something about $t(n,0)$, for example that $t(n,0) + a^2 \ge 1$ for all $n$. I doubt that.
