I'm trying to understand a proof that the teacher gave. The course is Calculus I and we are just starting, topics like series and integration have not been covered yet. We started with linearization and then Taylor's polynomial.
Theorem: If a function $f$ is N-times differentiable in $a$, then it exists only one polynomial (and it's $P_n(f, a)$) that satisfies: \begin{equation*} f(x) = P_{n, a}(x) + r((x - a)^n) \end{equation*} Here, $P_{n, a}(x)$ is the Nth-order Taylor polynomial of $f(x)$ at $x = a$ and \begin{equation*} \lim_{x\to a} \tfrac{r((x - a)^n)}{(x - a)^n} = 0 \end{equation*} For $n = 1$ the theorem is already proved (Mean value theorem I think).
The problem is that I don't understand why (he said) in order to prove the theorem it's enough to prove this limit: \begin{equation*} \lim_{x\to a} \tfrac{r((x - a)^n)}{(x - a)^n} = 0 \end{equation*} Or: \begin{equation*} r_n(x) = f(x) - P_{n, a}(x) \end{equation*} Basically what the teacher did is:
1) He evaluate the function for $x = a$ and concluded $r_n(a) = 0$: \begin{equation*} f(a) = f(a) + \tfrac{f^{(1)}(a)(a - a)}{1!} + \tfrac{f^{(2)}(a)(a - a)^2}{2!} + \dots + \tfrac{f^{(n)}(a)(a - a)^n}{n!} + r((a - a)^n) \end{equation*} 2) Did the same as in step 1 but this time he evaluate the differentiated function (instead of the original function) for $x = a$. He concluded $r^{(1)}_n(a) = 0$.
3) Did the same as in step 2 but for $n$ higher than 1, and he concluded $r^{(n)}_n(a) = 0$. I think he showed that $f$ is N-times differentiable in $a$ and the remainder is zero.
4) Then with the help of L'Hôpital's rule he did:
$\lim_{x\to a} \tfrac{r_n(x)}{(x - a)^n} = \lim_{x\to a} \tfrac{r^{(1)}_n(x)}{n(x - a)^{n - 1}} = \lim_{x\to a} \tfrac{r^{(2)}_n(x)}{n(n - 1)(x - a)^{n - 2}} = \dots = \lim_{x\to a} \tfrac{r^{(n - 1)}_n(x)}{n!(x - a)}$
So: \begin{equation*} r^{(n)}_n(a) = r^{(n - 1)}_n(a) \end{equation*} With the derivative definition and the result of step 3 he did:
$r^{(n)}_n(a) = \lim_{x\to a} \tfrac{r^{(n - 1)}_n(x) - r^{(n - 1)}_n(a)}{x - a} = 0$
He knows that $r^{(n - 1)}_n(a) = 0$ (also step 3), so: \begin{equation*} \lim_{x\to a} \tfrac{r^{(n - 1)}_n(x)}{x - a} = 0 \end{equation*} 5) Finally, because of step 4:
$\lim_{x\to a} \tfrac{r_n(x)}{(x - a)^n} = \lim_{x\to a} \tfrac{r^{(n - 1)}_n(x)}{n!(x - a)} = \lim_{x\to a} \tfrac{1}{n!}\tfrac{r^{(n - 1)}_n(x)}{x - a} = \lim_{x\to a} \tfrac{1}{n!}r^{(n)}_n(a) = 0$
I follow the proof, it's easy, but as I said I don't understand why in order to prove the theorem it's enough to prove the limit. I've searched on the Internet, read books (like Calculus, Vol. 1 by Tom M. Apostol) and sincerely I didn't found a proof like the one the teacher gave.