Consider the following proposition. $P_n(x)= \sum_{k=0}^{n} a_k (x-a)^k $ satisfies :
$\lim_{x\to a} \frac{f(x)-P_n(x)}{(x-a)^n}=0 $
if and only if $P_n$ is the $n$th Taylor polynomial of $f$ around $a$. I can prove the direction in which we assume $P_n$ is the Taylor polynomial, however the other direction (i.e. showing this is unique) has eluded me. Any help is appreciated.
Suppose that $P $ and $Q $ are degree $n $ polynomials with $$\lim_{x\to a} \frac{f(x)-P(x)}{(x-a)^n}=0,\ \ \ \lim_{x\to a} \frac{f(x)-Q(x)}{(x-a)^n}=0.$$ It follows that $$\tag {*}\lim_{x\to a}\frac {P (x)-Q (x)}{(x-a)^n}=\lim_{x\to a} \frac{f(x)-Q(x)}{(x-a)^n}-\lim_{x\to a} \frac{f(x)-P(x)}{(x-a)^n}=0. $$ Any polynomial can be written as its Taylor polynomial around $a $. So we may assume $$P (x)=\sum_{k=0}^na_k (x-a)^k,\ \ \ \ \ Q (x)=\sum_{k=0}^nb_k (x-a)^k. $$Now $(*) $ becomes $$\tag{**}0=\lim_{x\to a}\sum_{k=0}^n (a_k-b_k)(x-a)^{k-n}. $$ By multiplying by different powers of $x-a $ we obtain $a_k=b_k $ for all $k $.
Concretely: the first term in the sum in $(**)$ is $a_0-b_0$, so if the limit is zero we get that $a_0=b_0$, Now $(**)$ becomes, after multiplying by $(x-a)^{n-1}$, $$ 0=\lim_{x\to a}\sum_{k=1}^{n} (a_k-b_k)(x-a)^{k-1}. $$ The first term in the sum is $a_1-b_1$, so if the limit is zero then $a_1=b_1$. This reasoning can keep going until we get that $a_k=b_k$ for all $k$.