Prove $P_n(x) = \sum_{k=0}^{n}b_k(x-b)^k$ satisfies $$\lim\limits_{x\to b}\dfrac{f(x)-P_n(x)}{(x-b)^n} = 0 $$ iff $P_n = T_n$ Where $T_n$ is the nth Taylor polynomial of $f$ about $b$.
I know from Taylor's Theorem that $$\lim\limits_{x\to b}\dfrac{f(x)-T_n(x,b)}{(x-b)^n} = 0$$ I can see the forward proof intuitively from Taylor's theorem, but intuition isn't reliable.
Hint: Assuming $f$ is $n+1$-th continuously differentiable, then \begin{align} f(x)=&\ f(b)+f'(b)(x-b)+\ldots+ \frac{f^{(n)}(b)}{n!}(x-b)^n+\frac{f^{(n+1)}(\xi(x))}{(n+1)!}(x-b)^{n+1}\\ =&\ T_n(x)+\frac{f^{(n+1)}(\xi(x))}{(n+1)!}(x-b)^{n+1} \end{align} where $b<\xi(x)<x$. By the hypothesis, we have \begin{align} \frac{f(x)-P_n(x)}{(x-b)^n} = \frac{T_n(x)-P_n(x)}{(x-b)^n}+\frac{f^{(n+1)}(\xi(x))}{(n+1)!}(x-b). \end{align} Since the left hand side tends to zero as $x\rightarrow b$ then so does the right hand side. In particular, since \begin{align} \left\|\frac{f^{(n+1)}(\xi(x))}{(n+1)!}(x-b)^{n+1}\right\|\leq M|x-b|\rightarrow 0 \end{align} as $x\rightarrow b$, then \begin{align} \frac{T_n(x)-P_n(x)}{(x-b)^n} \rightarrow 0 \end{align} as $x\rightarrow b$.