Why is derivative of a function at a point the slope of the tangent at that point?

Question

Consider a function y = f(x).
Let one point A on the curve be (a , f(a)) and a second point B (a+h , f(a+h)).
$m_{AB} = (f(a+h) - f(a))/ h$.
As h → 0 point A and B come closer.
My question is:
As points A and B come closer, why does the line joining them have to be a tangent? Since the points are so close can't it be a chord passing through A? If we consider A and B as a single point, there are many lines passing through A.Then why a tangent?
Can anybody help?

Answer 1

You are right. The line $AB$ is a chord (or secant). Following up @sasquires comment, what you have to imagine is that there are two lines.

The first is the tangent at $A$. We are interested in finding its gradient, but we only have one point, so we cannot find it diectly.

The second line is the chord $AB$. We have two points so we can find its gradient.

Now imagine point $B$ moving closer to point $A$. The closer it gets, the closer its gradient gets to the gradient we are interested in (the gradient of the tangent at $A$). And if we make the distance between $A$ and $B$ infinitesimally small (but not zero), we can say that the gradient $AB$ is the same (almost) as the gradient of the tangent at $A$.

In fact, as $h$ approaches 0, the gradient $AB$ approaches the gradient of the tangent at $A$.