Differentiability and Local Linearity Intuitively I know that it is true that differentiability is equivalent to 'local linearity': When you zoom in on a differentiable function, it appears to be linear.
However, how can one formalizes this idea and prove it? Or is it more obvious from the definition than I thought?
 A: The usual way to define differentiability for functions $f:\mathbb{R}\to \mathbb{R}$ is that it is differentiable at point $a$ if the following limit exists:
$$\lim_{h\to 0} \frac{f(a+h)-f(a)}{h}$$
And then $f'(a)$ is defined to be equal to that limit. But there is an equivalent way to define differentiability which gives more insight (as it seems to me) into why differentiability is intuitively "local linearity". Let's do some rearangements. Suppose we have function $f$ which is differentiable at some point $a$ and its derivative there is equal to $k$. Then:
$$\lim_{h\to 0} \frac{f(a+h)-f(a)}{h}=k \quad\quad (*)$$
Let's subtract $k$ from both sides to get:
$$\lim_{h\to 0} \frac{f(a+h)-f(a)-k\cdot h}{h}=0$$
Let's look at the function $g(h)=f(a+h)-f(a)-k\cdot h$. What the limit above says is that:
$$\lim_{h\to 0} \frac{g(h)}{h}=0 \quad\quad (**)$$
From the defintion of $g(h)$ we can say that $$f(a+h)-f(a)=k\cdot h + g(h) \quad\quad(***)$$
What $(**)$ says can be intuitively interpreted as "$g(h)$ is really small compared to $h$ near $0$". But then $(***)$ basically means that "$f$ behaves just like linear function $k\cdot h$ near $a$" (neglecting the term $g(h)$). And this is why we can intuitively say "differentiability is local linearity".
So now if we want to make it into a rigorous definition, we define function $f$ to be differentiable at $a$ if there exists such a number $k$ that function $g$ which obeys $(***)$ will obey the property $(*)$. Now this answer basically contains proof that if function is differentiable under the usual definition, then it will be differentiable under the second definition. It is not a hard exercise to prove that if function is differentiable under second defintion, then it will be differentiable under the usual defintion (and by the way, this number $k$ will be equal to the limit $(*)$, as is expected). So this means that these two defintions are equivalent.
