Take an interval of the form ${ [\delta, 2A-\delta] \subseteq \mathbb{R} _{\gt 0} }.$ So here ${ 0 \lt \delta \lt A }.$ Consider ${ \log (x) }$ on ${ [\delta, 2A-\delta] }.$

We’ll find a sequence ${ (P _n (x)) }$ of polynomials of degrees ${ \deg(P _n) \leq n }$ such that ${ \max _{x \in [\delta, 2A - \delta]} \vert \log(x) - P _n (x) \vert }$ is ${ \mathcal{O} (\frac{1}{n}) }.$

That is, we’ll specify a seq of polys ${ (P _n (x)) }$ with ${ \deg(P _n) \leq n }$ and a constant ${ C \gt 0 },$ such that ${ \vert \log(x) - P _n (x) \vert \leq \frac{C}{n} }$ for all ${ x \in [\delta, 2A-\delta] }$ and all but finitely many ${ n \in \mathbb{Z} _{\gt 0} }.$

Consider ${ \log (1+x) }$ on ${ (-1, \infty) }.$ Its Taylor polynomials about $0$ are ${ S _n (x) = x - \frac{x ^2}{2} + \ldots + (-1) ^{n-1} \frac{x ^n}{n} }.$ Writing ${ \log(1+x) = S _n (x) + R _n (x) }$ we have ${ R _n ‘ (x) = \frac{1}{1+x} - (1 - x + \ldots + (-1) ^{n-1} x ^{n-1}) }$ ${ = \frac{1-(1+(-1) ^{n-1} x ^n)}{1+x} }$ ${ = (-1) ^n \frac{x ^n}{1+x} }.$ So ${ R _n (x) = R _n (0) + \int _{0} ^{x} R’ _n (t) dt }$ ${ = (-1) ^n \int _{0} ^{x} \frac{t ^n}{1+t} dt }.$
When ${ x \in [0,1] },$ we have ${ \vert R _n (x) \vert }$ ${ = \int _{0} ^{x} \frac{t ^n}{1+t} dt }$ ${ \leq \int _{0} ^{x} \frac{t ^n}{1+0} dt }$ ${ = \frac{x ^{n+1}}{n+1} }$ ${ \leq \frac{1}{n+1} .}$ When ${ x \in [-1 + \Delta, 0) }$ for some ${ \Delta \in (0,1) },$ we have ${ \vert R _n (x) \vert = \left\vert \int _{0} ^{-\vert x \vert} \frac{t ^n}{1+t} dt \right\vert }$ ${ \overset{ {\color{purple}{s =(-t)}} }{=} \left\vert \int _{0} ^{\vert x \vert} \frac{(-s) ^n}{1-s} (-ds) \right\vert}$ ${ = \int _{0} ^{\vert x \vert} \frac{s ^n}{1-s} ds }$ ${ \leq \int _{0} ^{\vert x \vert} \frac{s ^n}{1-\vert x \vert} ds }$ ${ = \frac{ \vert x \vert ^{n+1}}{(1-\vert x \vert)(n+1)} }$ ${ \leq \frac{\vert x \vert ^{n+1}}{\Delta (n+1)} }$ ${ \leq \frac{1}{\Delta (n+1)} }.$
Combining the estimates, we have : If ${ \Delta \in (0,1) },$ then ${ \max _{x \in [-1 + \Delta, 1]} \vert R _n (x) \vert \leq \frac{1}{\Delta (n+1)}. }$

Back to the original problem. For ${ x \in [\delta, 2A-\delta] },$ we see ${ \log(x) = \log(A) + \log(1 + \frac{x-A}{A}) }$ ${ = \log(A) + S _n (\frac{x-A}{A}) + R _n (\frac{x-A}{A}) }.$
Further ${ \max _{x \in [\delta, 2A-\delta]} \vert R _n (\frac{x-A}{A}) \vert }$ ${ = \max _{\frac{x-A}{A} \in [-(1-\frac{\delta}{A}), 1-\frac{\delta}{A}]} \vert R _n (\frac{x-A}{A}) \vert }$ ${ = \max _{u \in [-(1-\frac{\delta}{A}), 1-\frac{\delta}{A}]} \vert R _n (u) \vert }$ ${ \leq \frac{A/{\delta}}{n+1}. }$

So ${ \max _{x \in [\delta, 2A-\delta]} \vert \log(x) - (\log(A) + S _{n-1} (\frac{x-A}{A}) ) \vert \leq \frac{A/{\delta}}{n}, }$ as needed.

In fact here the polynomials ${ P _n(x) := \log(A) + S _{n-1} (\frac{x-A}{A}) }$ have degrees ${ \deg(P _n) = (n-1). }$