How to prove that as one zooms in on a point of the graph of a differentiable function, it looks more and more like the tangent at the point? How can one prove that as one zooms in on a point of the graph of a differentiable function of one variable, the graph looks more and more like the tangent line at the point?
One can see this phenomenon with the aid of computer graph plotters and their zooming capabilities, but how can one really prove this?
Also, it is intuitively clear, but what does it mean mathematically "to look more and more like the tangent line"? I guess we first need a proper mathematical definition of this intuitive notion before we proceed to a proof of the original question, but I have no idea how to obtain one (maybe it considers the "error" made by this tangent approximation in comparison to that of other straight lines at the same point? I really don't know).
 A: Why not mimic the computer graph plotter idea, namely that after rounding off coordinates to the pixels on a screen, they form a straight line, and otherwise we could compute the standard deviation (via the sum of squares) from the tangent pixels to the graph pixels:

Let us make it simple and define the following screen: $[-b,b]\times(-\infty,\infty)$ pixels and the graph of $f$ translated such that the point of interest $(a,f(a))$ becomes centered at $(0,0)$ on the screen. Then we define a $k$-zoom as meaning that one pixel represents $1/10^k$ units on the graph. For each value of $x:=a+n/10^k$ and corresponding $y=f(x)$ we plot the point $$\newcommand{\round}{\operatorname{round}}P:=\left(\round(10^k(x-a)),\round(10^k(y-f(a))\right)$$ on the screen and compare point by point to the same type of plot of the tangent $$t(x)=f'(a)(x-a)+f(a)$$


Then what we have to prove is that the sum of squares of $y$-differences between the pixel plot of $f$ and $t$ converge to zero (as $k$ tends to $\infty$). Note that those differences can be upper bounded by differences of the form $|y_f-y_t|+1$ so this suggests that solving the problem without rounding off to integer pixel values will essentially give us the same metric of how well they match.
