I'm reading through the proof of chain rule in Calculus with Analytical Geometry by George F. Simmons(Second Edition). The "almost" correct proof goes as follows. Here, y is a differentiable function of u and u is a differentiable function of x.
$$ \frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta u} .\frac{\Delta u}{\Delta x} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta u} . \lim_{\Delta x \to 0} \frac{\Delta u}{\Delta x} = \frac{dy}{du} . \frac{du}{dx}$$
The text goes on to say that the above is invalid because it doesn't consider the case when $\Delta u$ is unchanged for a small positive change in $\Delta x$. I can see that if $u(x)$ is a constant function, this can happen, but we say in the theorem (rule) that $u$ is a function of $x$. In what other cases can this happen? It seems to me that if $u$ is differentiable, then it's also continuous. And for a continuous function that's not a constant, how can $\Delta u$ be 0, no matter how close $\Delta x$ approaches 0?
In the Wikipedia article, I saw an example of $u = x^2.\sin(\frac{1}{x})$, but $\sin(\frac{1}{x})$ is not continuous at $x = 0$.