Suppose $f(x)$ has $n^{th}$ derivative in interval $(a,b)$ including $x_0$.
The Taylor series of $f(x)$ is $p_n(x)=f(x_0)+f'(x_0)(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2+...+\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n$
Prove that $\underset{x \to x_0}\lim\frac{f(x)-p_n(x)}{(x-x_0)^n}=0$.
Known:
Take $R_n(x)=f(x)-p_n(x)$, we have
$R_n(x_0)=R_n'(x_0)=...=R_n^{(n)}(x_0)=0$
Then we use L'Hoptial Rule n-1 times to compute the limit as
$\underset{x \to x_0}\lim\frac{R_n(x)}{(x-x_0)^n}=\underset{x \to x_0}\lim\frac{R_n'(x)}{n(x-x_0)^{n-1}}=...=\underset{x \to x_0}\lim\frac{R_n^{(n-1)}(x)}{n!(x-x_0)}=\frac{1}{n!}\underset{x \to x_0}\lim\frac{R_n^{(n-1)}(x)-R_n^{(n-1)}(x_0)}{x-x_0}=\frac{1}{n!}R_n^{(n)}(x)=0$
Question:
What I want to know is why cant't the L'Hopital Rule be used again in the last step, which is $\frac{1}{n!}\underset{x \to x_0}\lim\frac{R_n^{(n-1)}(x)-R_n^{(n-1)}(x_0)}{x-x_0}=\frac{1}{n!}R_n^{(n)}(x)$? It seems that both $R_n^{(n-1)}(x)$ and $x-x_0$ satisfy the requirement of L'Hopital Rule.
I see the proof in a textbook, but I don't really understand it. Thank you very much!
Welcome. Don’t get lost in L’Hopital - you could use it again, but that last limit is, by definition, the derivative of $R_n^{(n-1)}$ at $x_0$, $R_n^{(n)}(x_0)$. It all comes to the same thing.