Circular definition of tangent line and derivative I'm trying to understand the deep relations between the tangent line to the graph of a function $f$ at a given point $P$, and the derivative of $f$ at the same point.
Indeed, in many books the derivative is often defined as the slope of the "tangent", with only an intuitive definition of what a tangent line is. Then, once the concept of derivative is well assimilated, books define more precisely the tangent as... the line passing through $P = (a, f(a))$ with slope $f'(a)$ ! This seems a bit circular.
So some authors use an other approach, starting by giving a geometric definition of a tangent, and then showing the connection with derivatives. That's what one can read in the paper "What a tangent line is when it isn't a limit", by Irl Bivens. But I am missing something in the proof of Theorem 1, when the author writes:

it suffices to show that $|f(x)-L(x)|\leq (1/2)|L(x)-K(x)|$

where $f$ is a differentiable function, $L$ is the line of equation $L(x) = f'(a)(x-a)+f(a)$ and $K$ an other line, of equation $K(x)=p(x-a)+f(a)$ with $p \neq f'(a)$. So, basically, the goal is to show that $L$ is a better approximation of $f$ than any $K$.
But where does the right-hand side come from ? It looks like, locally, $f(x) = 1/2 (L(x)+K(x))$, but why?

Given that
$$
|L(x)-K(x)|=|L(x)-f(x)+f(x)-K(x)|\leq|f(x)-L(x)|+|f(x)-K(x)|\tag1,
$$
if 
$$
|L(x)-K(x)|\geq2|f(x)-L(x)|\tag2
$$
then
$$
2|f(x)-L(x)|\leq|f(x)-L(x)|+|f(x)-K(x)|,
$$
from which you obtain easily that
$$
|f(x)-L(x)|\leq|f(x)-K(x)|\tag3.
$$
