Why can you cancel out the dx in u-substitution?

[lugita15]

This is where choice is used. You can never give exact definition of the $x_n$. This is why it requires choice.

Hmm, color me skeptical. Why can't you explicitly construct a sequence satisfying $x_n \in (x - \frac{1}{n}, x + \frac{1}{n})$? For instance, $x_n$ can be the point a third of the way into the interval $(x - \frac{1}{n}, x + \frac{1}{n})$.

 

[micromass]

Hmm, color me skeptical. Why can't you explicitly construct a sequence satisfying $x_n \in (x - \frac{1}{n}, x + \frac{1}{n})$? For instance, $x_n$ can be the point a third of the way into the interval $(x - \frac{1}{n}, x + \frac{1}{n})$.

Note that you also want $|f(x_n)-f(x)|\geq \varepsilon$.