I am refreshing my calculus knowledge. One of the textbooks I use states what you can see down below (see the screenshot attached).
The question I have: must not the definition explicitly require $f$ to be continuous? Seems I am allowed to define $f$ as following:
$$ f(x) = \begin{cases} 0 &, x = 0 \\ 1 &, x \neq 0 \end{cases} $$
...so $f(0 + dx) = 1$ since $dx > 0$ by definition and thus $0 + dx \neq 0$, which, in turn, implies that $f(0 + dx) - f(0) = 1 - 0 = 1 \neq dy$.
Am I missing something important/obvious here?
You are right, $f(0+dx) - f(0) = 1$, but to take derivative, function has to be continuous (at the neighbourhood of the point, you are doing derivative at). If you do limit definition of derivative, you'll get limit of $\frac{1}{x}$, as $x$ approaches $0$, giving $\frac{1}{0}$, which is undefined.