Exercise 1.6.2 (Everywhere differentiable function not necessarily continuously differentiable)

Consider the function that vanishes quickly at $0$, yet also oscillates rapidly near that point:

Notice that on the domain $[-1,1]∕\{0\}$ both functions $\sin <!--\; FUNCTION\; APPLICATION\; -->(1∕x)$ and $x$ are differentiable (cf. Analysis II, section 4.7), and so is their composition and product (cf. Analysis I, Theorem 10.1.13 and 10.1.15). Thus, $f$ is differentiable on $[-1,1]∕\{0\}$ with derivative (see derivative rules)

What about the point $0$? Using raw machinery, we calculate the value of the derivative at this point:

But $\underset{x\to 0}{\lim <!--\; FUNCTION\; APPLICATION\; -->}{f}^{\prime }(x)$ diverges, since $\sin <!--\; FUNCTION\; APPLICATION\; -->(1∕x)$ part of it is periodic. Thus, $\underset{x\to 0}{\lim <!--\; FUNCTION\; APPLICATION\; -->}{f}^{\prime }(x)\ne f^{\prime }(0)$, and $f^{\prime }$ is not continuous.