# Is the "derivative limit theorem" true in the higher dimension?

~~I am asking yourself if the adhering to declaration is an approved theory in real analysis. Does any person below recognize the specific reference for it? It possibly an effect of the mean value theory, I assume. Encouraged by the answer to this inquiry, I interested concerning that if this declaration is still real in the greater measurement instance. ~~

The adhering to declaration is an effect of the Mean Value Theorem.

Allow $x_0\in{\bf R}$ and also $U(x_0)$ be the area of $x_0$.

$$f:U(x_0)\to{\bf R}$$

is continual on $U(x_0)$ and also differentiable on $U(x_0)\setminus\{x_0\}$. If the restriction

$$\lim_{x\to x_0;x\in U(x_0)\setminus\{x_0\}}f'(x)$$

exists, after that $f$ is differentiable at $x_0$ and also

$$f'(x_0)=\lim_{x\to x_0;x\in U(x_0)\setminus\{x_0\}}f'(x)$$

Here is my inquiry:

Is this declaration still real for the **greater measurement ** if $$f:U(x_0)\subset{\mathbb R}^n\to{\mathbb R}^m ?$$

Since I do not recognize any kind of generalization of the MVT in the greater measurement, I assume one might require to approach it straight necessarily. On the various other hand, if one can manage it "component - wise", one might have the ability to lower it to the one measurement instance.

It is a crucial attribute of modern-day multivariate calculus that it can and also need to be done - free.

We might think that $x_0=f(x_0)=\lim_{x\to 0}f'(x)=0$ and also need to confirm that $f'(0)$ exists and also is $=0$, that is to claim: For any kind of $\epsilon>0$ there is a $\delta>0$ with $|f(x)|<\epsilon|x|$ as quickly as $|x|<\delta$.

For the evidence we need to consider the mean - value theory for features $\phi: [a,b]\to{\mathbb R}$, due to the fact that in this instance it suffices that $\phi$ is differentiable in the indoor $]a,b[$ and also continual on every one of $[a,b]$.

So allow an $\epsilon>0$ be offered. By presumption there is a $\delta>0$ such that $\|f'(x)\|<\epsilon$ for all $x$ with $0<|x|<\delta$. ${\it Fix}$ such an $x$. If $f(x)=0$ there is absolutely nothing to confirm. Or else placed $u:=f(x)/|f(x)|$ and also take into consideration the supporting function $$\phi(t):=u\cdot f(t x)\quad (0\leq t\leq 1)\ .$$ Then by the chain regulation we get $$|f(x)|=|\phi(1)-\phi(0)|=|\phi'(\tau)|=|u\cdot (f'(\tau x).x)|\leq \|f'(\tau x)\| |x|< \epsilon|x|\ ,$$ as called for.

It additionally adheres to from L'Hospital is regulation related to $\frac{f(x)-f(x_0)}{x-x_0}$.

**Included ** (to make clear for the ones that downvoted the solution)

Just to make clear a little, in fact in this instance making use of L'Hospital is actually similarly as making use of Patrick evidence implicitely.

L' H is confirmed making use of Cauchy is mean value Theorem (which incidentally does not request for $f$ not be constant). Yet in our instance, $g(x)=x-x_0$, and also therefore Cauchy is Mean Value Theorem is specifically the typical Mean Value Theorem.

Finally, if you evaluate Patrick is evidence he primarily confirms uusing the MVT the following: If $f(x), x-x_0$ are continual on $U$, differentiable on $U-\{ x_0 \}$ and also $\lim_{x-x_0} \frac{f'(x)}{1} =L$ after that $\lim_{x-x_0} \frac{f(x)-f(x_0)}{x-x_0} =L$.

While his evidence is less complicated and also much cleaner than merely making use of L' H, this is simply a certain instance on L' H

It is, actually, an effect of the mean value theory ; intending your area has an open period fixated $x_0$, call the restriction of $f'(c)$ to be $L$, take $x$ in this period ; therefore there exists $c$ such that $$ f(x)-f(x_0) = f'(c)(x-x_0) \quad \Rightarrow \quad \frac{f(x) - f(x_0)}{x-x_0} = f'(c) \to L(x_0) $$ due to the fact that $c \to x_0$ when $x \to x_0$ and also $f'(c) \to L(x_0) = f'(x_0)$. There is a means of seeing this in the greater dimensional instance ; in $\mathbb R^n$, make use of similarly of reasoning, yet as opposed to calling it the mean value theory, usage Taylor is Theorem which claims that there exists a $c \in [x_0, x] \overset{def}{=} \{ y \in \mathbb R^n \, | \, y = \lambda x_0 + (1-\lambda) x_0, \, \lambda \in [0,1] \}$ such that \begin{align*} f(x) - f(x_0) &= f'(c)(x-x_0) \quad \Longrightarrow \quad \frac{f(x) - f(x_0) - L(x_0)(x-x_0)}{\| x-x_0 \|} \\ &= (f'(c) - L(x_0)) \left( \frac{x-x_0}{\| x-x_0 \| } \right) \to 0. \end{align*} My debates are a little questionable ; I'm simply offering you a suggestion of what it resembles. You require to beware what takes place in greater measurement to what takes place when you review that the last term mosts likely to $0$.