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.


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 $x_0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.