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

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In the discussion about canceling out dx in u-substitution, participants clarify that the notation du/dx represents the derivative as a limit, not a fraction, which complicates the idea of cancellation. The integral notation dx serves as a reminder of the connection to Riemann sums, and while it may seem intuitive to cancel dx, this requires a deeper understanding of calculus principles, particularly the chain rule and the fundamental theorem of calculus. The conversation also touches on the validity of infinitesimals in mathematics, noting that while some argue against their use, others, like Abraham Robinson, have established a rigorous framework for them. Ultimately, the discussion emphasizes that the process of u-substitution is not merely a trick but involves significant mathematical reasoning and definitions. Understanding these concepts is crucial for grasping the underlying mechanics of integration and differentiation.
  • #31
micromass said:
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}).
 
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  • #32
lugita15 said:
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.
 

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