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Let $f(x)$ be a real-valued function defined on a closed interval [a, b], differentiable on the open interval (a, b) $n-1$ times.

$x_0$ belongs to [a, b]. Suppose that we have:$$f(x_0)=0,f'(x_0)=0,f''(x_0)=0,...,f^{(n-1)}(x_0)=0$$ Also suppose that the function admits the nth derivative at $x_0$. In this case my book says that is:$$f(x)=\frac{(x-x_0)^n}{n!}[f^{(n)}(x_0)+\sigma(x)]$$ where:$$\begin{matrix} \lim_{x \to x_0}\sigma(x)=0 \end{matrix}$$ It seems that in my book's proof must be (because it isn't completely clear ): $$\begin{matrix} \lim_{x \to x_0}f^{(n)}(x)\end{matrix}=f^{(n)}(x_0)$$ It means that $f^{(n)}(x)$ is continuous at $x=x_0$...but in the theorem it isn't mentioned that $f^{(n)}(x)$ is continuous at $x=x_0$; I know that the first $(n-1)$th derivatives are continuous in $x_0$ and $f^{(n)}(x_0)$ exists. So is the continuity of $f^{(n)}(x)$ at $x=x_0$ necessary to prove the theorem?

-Proof's book

If $x\ne x_0 $ and $n>1$, according to Cauchy's mean value theorem, we have:$$\frac{f(x)}{\phi(x)}=\frac{f^{(n-1)}(\eta)}{\phi^{(n-1)}(\eta)}$$

where $\eta$ is a suitable interior point of the interval $[x_0,x]$ and $\phi(x)=(x-x_0)^n$. Since:$$\phi^{(n-1)}(\eta)=n!(\eta-x_0)$$$$f^{(n-1)}(x_0)=0$$the last formula I wrote becomes:$$\frac{f(x)}{(x-x_0)^n}=\frac{1}{n!}\cdot\frac{f^{(n-1)}(\eta)-f^{(n-1)}(x_0)}{\eta-x_0}$$ Now: $$\begin{matrix} \lim_{x \to x_0}(n!\frac{f(x)}{(x-x_0)^n}) \end{matrix}=\begin{matrix} \lim_{x \to x_0}\frac{f^{(n-1)}(\eta)-f^{(n-1)}(x_0)}{\eta-x_0} \end{matrix}=f^{(n)}(x_0)$$I'm stuck here: in the second limit; I think that writing means that $f^{(n)}(x)$ is continuous at $x=x_0$. Am I right? Let's keep on with the proof; let:$$\sigma(x)=n!\frac{f(x)}{(x-x_0)^n}-f^{(n)}(x_0)$$ for $x\ne x_0 $ and $\sigma(x_0)=0$ we have: $$\begin{matrix} \lim_{x \to x_0}\sigma(x) \end{matrix}=0$$ Finally:$$f(x)=\frac{(x-x_0)^n}{n!}[f^{(n)}(x_0)+\sigma(x)]$$QED

My doubt is: is the continuity of $f^{(n)}(x)$ at $x=x_0$ necessary to prove the theorem? If it isn't can someone explain me why the second limit is valid?

Any help would be greatly appreciated.

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No, the proof doesn't imply that $f^{(n)}$ is continuous.

First, note that $\eta$ is in $[x,x_0]$, and depends on $x$. So one can write it like $\eta(x)$ and notice that $\eta(x) \xrightarrow{x \rightarrow x_0} x_0$, such that if the following limit exist

$$\lim_{t \rightarrow x_0}\frac{f^{(n-1)}(t) - f^{(n-1)}(x_0)}{t - x_0}$$

it must be equal to the original one by composition. But the previous limit is the growth rate, which is precisely the definition of the derivative of $f^{(n-1)}$ at point $x_0$. It doesn't require $f^{(n)}$ being continuous, but just defined.

Saying that $f^{(n)}$ is continuous is a stronger property, which is that $f^{(n)}(x) \xrightarrow{x \rightarrow x_0} f^{(n)}(x_0)$. It says that not only the growth rate can be defined, but also that its value varies continuously with $x_0$.

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