TITLE: Rigorously prove that the tangent line is indeed tangent?
QUESTION [5 upvotes]: Let $f$ be a function of $x$ and $f'$ be the derivative, at a point $(x_0, f(x_0))$
the slope is $f'(x_0)$, we know from any calculus book that
the line $g(x) = f(x_0) + f'(x_0)(x - x_0)$ is tangent to the curve of $f$, 
but how do you rigorously prove it?

REPLY [2 votes]: This is a nice question. There is a geometrical notion of a tangent line, and various definitions that one could choose in order to embody that idea. One then wants to show that this matches up with one of the definitions from calculus.
A couple of definitions from calculus:
(C1) Define the tangent line at P as the line that passes through P and has the same slope as the derivative. This doesn't work in cases where the line is vertical at P.
(C2) A better calculus-based definition is to use a parametrized curve, and define the tangent line as the one that points in the direction of $(\dot{x},\dot{y})$. This works when the tangent line is vertical.
Geometrically, several pretty good possibilities have been proposed:
(G1) (Apollonius) Given a point Q not on the curve, let P be the point on the curve closest to Q. Then the tangent line at P is the line through P that is perpendicular to PQ.
(G2) (Descartes) Let P be a point on the curve, and let C be a circle that touches the curve only at P. Since Euclid has already defined tangency to a circle (a tangent line touches the circle only at one point), we take the definition of a tangent to the curve and reduce it to the definition of the tangent to C.
(G3) Let P be a point on a curve. A tangent line at P is one such that if we rotate the line clockwise about P by an arbitrarily small angle, then the rotated line cuts across the curve at P, and if we rotate counterclockwise by an arbitrarily small angle, we cut across the curve in the opposite direction.
Of G1, G2, and G3, none works at the end-point of a graph.
It seems odd to me that although G1 and G2 are obviously very closely related, if not equivalent, Descartes is credited with G2, even though he lived a thousand years after Apollonius. Maybe Descartes progressed further in applying the definition.
I seem to remember seeing G3 attributed to an ancient Greek mathematician by someone on SE, but I can't find it.
A definition similar to G3 is given in Marsden and Weinstein, Calculus Unlimited, http://resolver.caltech.edu/CaltechBOOK:1981.001 . They do all the work of connecting this definition to C1. They don't handle the case of a vertical tangent line. It's quite straightforward, for example, to apply G3 to the parabola $y=x^2$ at $x=1$.
It's not obvious to me whether there is any meaningful sense in which one could really prove one the most general calculus-based definitions, such as C2, to be equivalent to one of these geometrical definitions. You'd have to deal with pathological cases like $y=\sin (1/x)$ at $x=0$, for which C2 gives a vertical tangent line (but fails to define the orientation of the line). Classical geometry never dealt with this kind of thing. For this example, I think G1 and G2 fail, while I'm not sure if G3 can be applied.