MHB How Does Definition 8.9 Imply Differentiability Near Point p?

[Math Amateur]

I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

I am currently reading Chapter 8: Differentiable Maps and am specifically focused on Section 8.2 Differentials ...

I need help in fully understanding Browder's comments on Definition 8.9 ... ...

Definition 8.9 including Browder's remarks reads as follows:View attachment 7469
In the above text from Browder, we read the following:

" ... ... We say that $$T \ : \ \mathbb{R}^n \mapsto \mathbb{R}^m$$ is an affine map if it has the form $$x \mapsto c + Lx$$ for some $$c \in \mathbb{R}^m$$ and linear map $$L \ : \ \mathbb{R}^n \mapsto \mathbb{R}^m$$. Thus Definition 8.9 says roughly that a function is differentiable at $$p$$ if it can be approximated near $$p$$ by an affine function. ... ..."I cannot see exactly how Definition 8.9 implies that a function is differentiable at $$p$$ if it can be approximated near $$p$$ by an affine function ... that is a function of the form $$x \mapsto c + Lx$$ ...

Can someone please demonstrate rigorously that Definition 8.9 implies that a function is differentiable at $$p$$ if it can be approximated near $$p$$ by an affine function ... that is a function of the form $$x \mapsto c + Lx$$ ... ?
Help will be much appreciated ...

Peter

 

[HallsofIvy]

It's a direct calculation. "If function F(x) can be approximated by an affine function near p" then there exist c and L such that F(x)= c+ L(x- p)+ \epsilon(x-p) where \epsilon(x-p) is the "error" in the approximation that goex to 0 as x goes to p (\epsilon(0)= 0).

Then F(p)= c+ \epsilon(0)= c and F(p+ h)= c+ L(h)+ \epsilon(h).
So F(p+ h)- F(p)= Lh+ \epsilon(h)

\frac{1}{|h|}\left(F(x+p)- F(p)- Lh\right)= \frac{1}{|h|}\left(Lh+ \epsilon(h)\right)= L\frac{h}{|h|}+ \frac{\epsilon}{|h|}.

We only need that we can choose \epsilon(x) goes to 0 faster than x.