Showing that the remainder term in Taylor's Theorem Converges to Zero

Question

On pg. 110 of Rudin's Principles of Mathematical Analysis, it is shown that if $f$ is a real function on $[a, b]$ with $f^{(n)}(t)$ existing for every $t \in (a,b)$, then there exists some $x \in (a, b)$ such that

$$ f(b) = \sum_{k=0}^{n-1} \frac{f^{(k)}(b)}{k!} (b - a)^k + \frac{f^{(n)}(x)}{n!} (b - a)^n $$

Now on wikipedia, Taylor's Theorem is stated at least once as the following:

$$ f(b) = f(a) + f'(a)(b-a) + \ldots + h_n(b)(b-a)^n $$

with

$$ \lim_{b \to a} h_n(b) = 0 $$

Question: Combining these two proofs/statements, it seems that

$$ \lim_{b \to a} h_n(b) = \lim_{b \to a} \frac{f^{(n)}(x_b)}{n!} = 0 $$

where $x_b$ is chosen for each $b$ as it converges to $a$.

But why? That is, how do we know that the error term that Rudin derives converges to zero, not as $n$ goes to infinity, but as $x \rightarrow a$?

Answer 1

We don't know that the error term goes to zero as $x\to a$. There are functions which are infinitely differentiable on $(0,1)$ but are not analytic at $0$. For example, if you plug in $$f(x)=\sin\left(\frac1x\right)$$ into your statements, takeing $f(0) = 0$, you will find that Rubin's statement, with $a = 0$ and arbitrary $n$ and $b$, holds, that the Taylor series converges with $n$ for all $x$ in $(0,1)$, but that the error term does not converge to zero as $b$ approaches zero.

Answer 2

I believe you misread what Wikipedia stated. It states that $$ f(b) = f(a) + f'(a)(b-a) + \ldots + \frac{f^{(n)}(a)}{n!}(b-a)^n +h_n(b)(b-a)^n. $$ So comparing the two notations, we get $$h_n(b)= \frac{f^{(n)}(x_b)}{n!}-\frac{f^{(n)}(a)}{n!}. $$ Does this alleviate your confusion? If not, let me know and I will edit/expand.

Answer 3

First, you need to note that the theorem comes in various forms, with various hypotheses. This is relevant for a few reasons. First, the hypotheses in the two versions you cite are different. Second, the result you claim Rudin states is obviously false. I'd bet a huge sum that the statement in the book is correct and you copied it incorrectly.

So you should get the hypotheses in Rudin's version straight. And you need to learn the hypotheses in that other version. Turns out that the hypotheses in Rudin's version are much stronger, and the conclusion is also much stronger. So the conclusion Rudin's theorem should imply that of Peano's, and it does; what @Aweygan said is correct, and more or less obvious once you get the theorems straight.