TITLE: What does "finite but unbounded" mean?
QUESTION [10 upvotes]: In the following example, Gelbaum and Olmsted (Counterexamples in Analysis, 1964,  p.37) speak of a derivative being “finite but unbounded”.

A function whose derivative is finite but unbounded on a closed interval.
The function
$$f(x)=\begin{cases}x^2\sin\dfrac1{x^2}&\text{if }x\neq 0\\
0&\text{if }x= 0\end{cases}$$
has the derivative
$$f'(x)=\begin{cases}2x\sin\dfrac1{x^2}-\dfrac2x\cos\dfrac1{x^2}&\text{if }x\neq 0\\
0&\text{if }x= 0,\end{cases}$$
which is unbounded on $[-1,1]$.

What does this mean and how is it possible that a quantity is both finite and unbounded?

REPLY [3 votes]: The theorem indicates that the derivative  of f(x) is finite at $x=0$ and it is unbounded on $[-1,1]$
It is obvious that a finite set is bounded, therefore the author's intention is  to show a function which has a finite derivative at $x=0$ but unbounded derivative on the interval [0,1].