Understanding why $0 = \lim_{x \to x0} \frac{\sum_{k=0}^{n} a_k (x - x_0)^k}{(x - x_0)^n}$ means $a_k = 0$ in a proof of Taylor's Theorem I was perusing an answer to this question here, but I'm confused regarding the last line
$0 = \lim_{x \to x0}  \frac{\sum_{k=0}^{n} a_k (x - x_0)^k}{(x - x_0)^n}$ 
this implies that $a_0 = a_1 = ... = a_n = 0$. I'm not sure why it implies that the coefficients are equal to  zero. I do see that the limit of the numerator must be zero, but that would only imply the first coefficient $a_0$ (when $k = 0$ ) must be zero, but for $k > 0
$, why do the coefficients have to be zero? My understanding is that as $x \to x_0$, $(x - x_0) \to 0$, then the coefficients can be any value?
 A: It may be easiest to pretend that $n = 1$ for a moment. Then the limit is
$$ \frac{a_0}{(x - x_0)} + a_1.$$As $x \to x_0$, the denominator in the first term blows up, so that if $a_0 \neq 0$ then there is no limit at all. So $a_0 = 0$. The last term is independent of $x$. But as we know the limit is $0$, we find that $a_1$ must also be zero.
It's the same basic reasoning for longer expansions: each denominator would blow up in the limit, forcing the corresponding numerators to be $0$; and thus the constant must also be $0$.
A: If
$$L=\lim_{x \to x_0} \sum^{n}_{k=0}\frac{a_k}{(x-x_0)^{n-k}}=0 $$
then every coefficient $a_k=0.$
If the unique non-zero coefficient is $a_n$ then the limit is limit of a constant $a_n$, so $a_n=0$.
Define the set $A=\{k\,|\,a_k \neq 0 \} $, if $A$ is empty then $a_k=0,\; \forall k$. Suppose $A$ is not empty, then it has a minimum $s<n$ (the case $s=n$ it's not possible by our first argument ) such that $a_s \neq 0$.
We have $\lim\limits_{x \to x_0} (x-x_0)^{n-s}=0$, and by the   multiplication of limits,
$$0= \lim_{x \to x_0} (x-x_0)^{n-s} \lim_{x \to x_0} \sum^{n}_{k=s}\frac{a_k}{(x-x_0)^{n-k}} $$
$$=\lim_{x \to x_0} \sum^{n-s}_{k=0}(x-x_0)^{n-s}\frac{a_{k+s}}{(x-x_0)^{n-s-k}}=\lim_{x \to x_0} \sum^{n-s}_{k=0}a_{k+s}(x-x_0)^{k}=a_s $$
so $a_s=0$ a contradiction. Then $a_k=0$ $\forall k.$
