Show that $a_0=a_1=...=a_n=0$ if $ \lim\limits_{x \rightarrow x_0} \frac{Q(x)}{(x-x_0)^n} =0$ 
Quoting: Let $Q(x) = a_0 +a_1(x-x_0)+...+a_n(x-x_0)^n$ be a polynomial of degree $\leq n$ such that 
  $$ \lim\limits_{x \rightarrow x_0} \frac{Q(x)}{(x-x_0)^n} =0$$ 
  Show that $a_0=a_1=...=a_n=0$

I understand that if $f^{(n)}(x_0)$ exists for $n \geq 1$, the limit theorem involving the n$^{th}$ Taylor Polynomial about $x_0$ is of the form 
$$
\lim\limits_{x \rightarrow x_0} \frac{f(x)-T_n(x)}{(x-x_0)^n}=0
$$
Considering that the form of $Q(x)$ corresponds to $T_n(x)$, and that it is assumed that $x \neq x_0$, we must have
$$
\lim\limits_{x \rightarrow x_0} f(x)-T_n(x)=\lim\limits_{x \rightarrow x_0} q(x)-Q(x) =0
$$As $q(x)$ is non-existing, It shows that $Q(x)= 0$. It follows that $a_0=a_1=...=a_n=0$
Would this be correct? Anything to rectify? Input is much appreciated.
 A: An easy way to prove that $Q(x)=0$ is by induction on $n$:


*

*For $n=1$ we have that $Q(x)=a_0+a_1(x-x_0)$ and the wanted limit is:
$$\lim_{x\to x_0}\frac{a_0+a_1(x-x_0)}{x-x_0}=0$$
Let $$g(x)=\frac{Q(x)}{x-x_0}\Rightarrow Q(x)=g(x)(x-x_0)$$
Now, since $\lim\limits_{x\to x_0}g(x)=0$ and $Q$ is continuous as a polynomial, we have:
$$Q(x_0)=\lim_{x\to x_0}Q(x)=\lim_{x\to x_0}g(x)(x-x_0)=0\cdot0=0$$
So, $$Q(x_0)=0\Rightarrow a_0+a_1(x_0-x_0)=0\Rightarrow a_0=0$$
and our limit now is:
$$0=\lim_{x\to x_0}\frac{a_1(x-x_0)}{x-x_0}=a_1\lim_{x\to x_0)}\frac{x-x_0}{x-x_0}=a_1\cdot1=a_1\Rightarrow a_1=0$$
So, $$Q(x)\equiv0$$

*Now, let us suppose that the requested is true for $n$, so that
$$\lim_{x\to x_0}\frac{P(x)}{(x-x_0)^n}=$$
for a polynomial $P$ of degree $\leq n$ implies that $P(x)\equiv0$. Let $$Q(x)=a_0+a_1(x-x_0)+\dots+a_{n+1}(x-x_0)^{n+1}$$
and let 
$$\lim_{x\to x_0}\frac{Q(x)}{(x-x_0)^{n+1}}=0$$
We will at first show that $a_0=0$. Let $$g(x)=\frac{Q(x)}{(x-x_0)^{n+1}}\Rightarrow Q(x)=g(x)(x-x_0)^{n+1}$$
So, since $\lim\limits_{x\to x_0}g(x)=0$ and $Q$ is continuous, we have:
$$Q(x_0)=\lim_{x\to x_0}Q(x)=\lim_{x\to x_0}g(x)(x-x_0)^{n+1}=0$$
So, $Q(x_0)=0\Rightarrow a_0=0$
Then $$\begin{align*}Q(x)=&a_1(x-x_0)+\dots+a_{n+1}(x-x_0)^{n+1}=\\=&(x-x_0)\underbrace{\left(a_1+a_1(x-x_0)+\dots+a_{n+1}(x-x_0)^n\right)}_{P(x)}=\\
=&(x-x_0)P(x)
\end{align*}$$
where $P(x)$ is a polynomial of degree $\leq n$. Now, our limit is:
$$\lim_{x\to x_0}\frac{Q(x)}{(x-x_0)^{n+1}}=\lim_{x\to x_0}\frac{(x-x_0)P(x)}{(x-x_0)^{n+1}}=\lim_{x\to x_0}\frac{P(x)}{(x-x_0)^n}=0$$
but, due to our hypothesis, since $P$ is of degree $\leq n$, we have that:
$$P(x)\equiv0$$
So:
$$a_1=a_2=\dots=a_{n+1}=0$$
And the proof is now complete.

A: We don't really need induction here (or, at a pinch, finite induction).
Multiplying by  $(x-x_0)^{n-k}$, we deduce that $\dfrac{Q(x)}{(x-x_0)^k}\to0$ for all $k=0,1,\dots n$.
In particular, $Q(0)=0$, whence $a_0=0$.
Next $\dfrac{Q(x)}{x-x_0}=a_1+a_2(x-x_0)+\dots+a_n(x-x_0)^{n-1}\to 0$, so $a_1=0$.
The result follows looking successively at all $\dfrac{Q(x)}{(x-x_0)^k}$.
