0

I have been reading the 1st volume of Courant’s “Differential and Integral Calculus” (https://www.ime.usp.br/~gorodski/ps/Courant-DifferentialIntegralCalculusVolI.pdf) and I have entered into some confusion. The chapter I am currently on is about the continuity of functions, and the part I demand clarification for is the subchapter spanning the pages 51-54, namely “Points of discontinuity”. I cannot quite understand what are the exact principles the author uses in order to define those points. Is it necessary that the function is (non-)defined at those points? Perhaps, the function must be defined in its (the point’s) deleted neighbourhood? How do we classify the discontinuities when, say, the limit from the right is infinite, and the left one is finite (or does not exist)? I really need to clear this confusion, for without this I cannot advance further through the book. What I would really want you to do is to read the mentioned subchapter (using the link above) and explain it to me.

Thank you!

Barbatulka
  • 457
  • 2
  • 6
  • I don't think there are any exact principles which cover all cases; those that he mentions are just some cases that appear frequently in practice. – Hans Lundmark Sep 29 '21 at 06:53
  • Also note that there is some debate about whether it's actually appropriate to use the word “discontinuity” for points where the function is undefined; see this question and the questions linking to it, for example this one. – Hans Lundmark Sep 29 '21 at 06:53

1 Answers1

1

I skimmed the section, I did not read it. (That's a bit much to ask of folks on a free knowledge sharing site). It is not necessary that the function be undefined.
Places (points) where a function is discontinuous occur because either the function is not defined there, or it is but we cannot force our function to stay as close as we want to the y-value by restricting the x value to a small enough zone.

This can fail in several ways, only a few of them have special names to my knowledge. The first is removable discontinuity: This is a function who has a well defined limit at that point, but isn't defined at that point. In other words, there's a hole that we could just fill in the graph to make it continuous. Example:

$$f(x)=\frac{x-2} {x^2-4}$$ note it is not defined at $x=2$, but everywhere $x\neq 2$ we can factor the bottom into $(x-2)(x+2)$, so for every point OTHER than $x=2$, $f$ is identical to the graph of

$$f(x)=\frac 1 {x+2}$$

Thus, it would have a removable discontinuity at 2, that we could remove by defining an extension of $f$ to include $2$ in the domain, with the output the limit value, here $f(2)=\frac 1 4$

A second named kind of limit is called a "jump discontinuity." This occurs when we have a function have one-sided limits approaching the point, but the two don't agree, so the function suddenly jumps. There's no way to remove this.

For example, we could have the function that on $x\leq 0$, $f(x)=-1$ but for $f(x)\geq 0$, $f(x)=1$ (This is just the sgn function that gives the sgn of a function, though also defined at 0 whereas that point is undefined in the actual sgn function). Here at 0 we have a "jump".

Other ways of failing are when one or both sides grow with out bound, like $\frac 1 x$ does near $0$. Finally, a function can oscillate wildly approaching a point, so that no matter how small you zoom in near the x value you can't get the $y$ value to settle down to stay near a point. The classic example here is $$\sin (\frac 1 x)$$

Note that as $x\to 0$, $\frac 1 x$ grows wildly. Since $\sin$ is $2\pi$ periodic, this means that in ANY small neighborhood around 0, the function will take on all outputs of $\sin$, namely $[-1,1]$ and thus never settle down.

Alan
  • 16,582