What is the intuitive reason for a function to be continous but not differentiable

Question

I was just wondering if the following is the right idea for the function to be continuous, but is not differentiable. So, intuitively the function is continuous whenever if we get close to a point a, then f(x) gets close to f(a). If a function is differentiable at a, then we can calculate the tangent line at a. So in order to stop a function from being differentiable at a. We just have to have a function which is continuous whose tangent line oscillate really rapidly to stop it from being differentiable. Is that the right idea ?

Answer 1

No. The function is not differentiable at $a$ when there is no tangent line to the graph at the point $\bigl(a,f(a)\bigr)$. The most typical reason for non-existence of a tangent there is when there is a corner at that point. This has nothing to do with oscillations.

However, the oscillation phenomenon also occurs. That's what happens with the function$$\begin{array}{rccc}f\colon&\mathbb R&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}x\sin\left(\frac1x\right)&\text{ if }x\neq0\\0&\text{ if }x=0.\end{cases}\end{array}$$It is not differentiable at $0$, although it is continuous there, because the quotiente$$\frac{f(x)-f(0)}{x}$$keeps oscilatting between $1$ and $-1$ as $x$ approaches $0$.

Answer 2

As @José Carlos Santos pointed out. A function is not differentiable at $a$ when there is no tangent line to the graph at the point $(a,f(a))$. An example of this is $f(x)=|x|$ as you can see the problem at $0$ in this graph:

Weierstrass later combined this idea with fractals, to produce his monster, a function which is continuous everywhere but nowhere differentiable. For more see Weierstrass Monster Function.