Question: In [this](https://en.wikipedia.org/wiki/Taylor%27s_theorem#Proof_for_Taylor.27s_theorem_in_one_real_variable) link I don't understand two last lines in that segment. 

> 
> \begin{align}
> \lim\_{x\to a} \frac{f(x) - P(x)}{(x-a)^k} &= \lim\_{x\to a} \frac{\frac{d}{dx}(f(x) - P(x))}{\frac{d}{dx}(x-a)^k} = \cdots = \lim\_{x\to a} \frac{\frac{d^{k-1}}{dx^{k-1}}(f(x) - P(x))}{\frac{d^{k-1}}{dx^{k-1}}(x-a)^k}\\
> &=\frac{1}{k!}\lim\_{x\to a} \frac{f^{(k-1)}(x) - P^{(k-1)}(x)}{x-a}\\
> &=\frac{1}{k!}(f^{(k)}(a) - f^{(k)}(a)) = 0
> \end{align}
> 
> 
> 

I know it says that it follows from the definition of the derivative at $x=a$ but can someone explain this to me more thoroughly ?

Answer: 
First, remember that 
$$
 \lim\_{x\to a} \frac{g(x) - g(a)}{x-a} = g'(a)
$$
whenever $g$ is differentiable at $a$. Some people are more familiar with the limit in the form $\lim\_{h\to0} \frac{f(a+h) - f(a)}{h}$, but they are equivalent by changing the variables from $h$ to $x=a+h$. 

Also, there seems to be a typo in the Wiki article. It says $f^{(j)}(a) = P^{(j)}(a)$ for all $j$ from $1$ to $k-1$, but it also holds for $j=k$.

The rest of it is the old add-and-subtract-the-same-thing trick:
$$
\begin{split}
\lim\_{x\to a} \frac{f^{(k-1)}(x) - P^{(k-1)}(x)}{x-a}
&= \lim\_{x\to a} \frac{f^{(k-1)}(x) -f^{(k-1)}(a) + f^{(k-1)}(a) - P^{(k-1)}(x)}{x-a} \\
&= \lim\_{x\to a}\left(\frac{f^{(k-1)}(x) -f^{(k-1)}(a)}{x-a} - \frac{P^{(k-1)}(x) - f^{(k-1)}(a)}{x-a}\right) \\
&= \lim\_{x\to a}\frac{f^{(k-1)}(x) -f^{(k-1)}(a)}{x-a} - \lim\_{x\to a}\frac{P^{(k-1)}(x) - P^{(k-1)}(a)}{x-a} \\
&= f^{(k)}(a) - P^{(k)}(a) = f^{(k)}(a) - f^{(k)}(a) = 0 
\end{split}
$$
This follows from L'Hopital Rule applied $k$ times: [see here](https://en.m.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule). 
This isn't just an immediate consequence of the definition of derivative.