Question about Peano form of the remainder 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 $n&gt;1$, 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.
Proof's book:

If $x\ne x_0 $ and $n&gt;1$, 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.