Derivative axiom

Question

This is confusing me very much... Is there (rigorous) proof that slope of secant line "goes to" slope of tangent line on some point when $\Delta x \rightarrow 0$? This is actually not obvious at all.

Intuitive problem: What guarantees me that function will not be all messy at infinite small input changes? How one can prove that function output sensitivity can't be "good" enough to produce such curve ("output machine") that there is no such small $\Delta x$ for which $\text{slope(secant)} = \text{slope(tangent)}$? I thought about structure of $\mathbb{R}$ and series but I am stuck. Thanks

Answer 1

Have you seen the following example? $$ f(x) = \begin{cases} x^2 \sin\frac1x, & x \neq 0 \\ 0, & x = 0 \\ \end{cases} $$

It has $f'(0) = 0$ but it's not obvious what one should mean with a tangent at $x=0$ since the graph oscillates fast close to origin.

Answer 2

Well, it's not true in general, even if the function in question is continuous. A simple example is the absolute value function, where as we approach the origin from the left the slope of the secant line is always $1$ but as we approach the origin from the right the slope of the secant line is always $-1$.

A more interesting example comes from trigonometry: consider the function $$f(x)=\begin{cases}x\sin({1\over x}) & \mbox{ if } x\not=0,\\ 0 & \mbox{ otherwise.}\\ \end{cases}$$

This function is pretty nice at first glance - in particular, it's continuous (this is a good exercise) - but around $x=0$ the secant lines behave quite poorly: for every $a\in [-1,1]$ and every $\epsilon>0$ there is some $y\in (0, \epsilon)$ such that ${f(y)-0\over y-0}=a$.

The relevant term here is differentiability. The above $f$ is an example of a continuous function which is not always differentiable (namely, it is not differentiable at $x=0$). There are even continuous functions which are nowhere differentiable!