Talk:Calculus/Continuity

Equivalent definition
$$\lim_{h \rightarrow 0} f(c+h) = f(c)$$, worth mentioning.


 * Or for negative one-sided continuity, $$\lim_{h \rightarrow 0} f(c-h) = f(c)$$.

Q and A
I'm just learning, but I have a question--in point #1 for the definition of continuity, shouldn't that be "that x is defined at c, so f(c) exists" rather "than that f is defined at c, so f(c) exists?"--anon 1


 * No, what is written is correct. You need that the function f is defined at c so that f(c) makes sense. Juliusross 12:45, 31 October 2005 (UTC)

Also, I changed the formatting, made it a list as I found it difficult to read how it was--change it back if you think that was bad. I didn't change any content though.--anon 1

It seemed odd that the definition of continuity didn't include all three conditions needed for continuity, so I added the other two. If someone could make it look pretty that would be great. On a completely different note, it would also be good if we could get an example of all three conditions failing, as well as some pictures?--anon 1

Yes pictures are both important and necessary.--Cronholm144 10:23, 19 June 2007 (UTC)

I reverted anon one, the conditions you added are necessary but not sufficient conditions of continuity. i.e. The definition already implies the conditions.--Cronholm144 10:28, 19 June 2007 (UTC)

Expanding
One of the ways this can be expanded is with the intermediate value theorem (IVM). Zginder 14:06, 8 November 2007 (UTC)

Inconsistency
We currently state that:
 * Remember that, if f is built out of rational, trigonometric, exponential and logarithmic functions, then the limit of f at any point in its domain is just its value at that point. So, by our definition, all such functions are continuous.

Which is certainly one point of view. My first comment would be that telling students tan(x) or 1/x are continuous because where they are discontinuous they are not defined misses the spirit of things a bit. A more serious problem is our first example of a removable discontinuity is $$x^2-9 \over x-3 $$, which is certainly a rational function, so we just finished saying it was continuous. Thenub314 (talk) 18:57, 23 June 2009 (UTC)

This is a more or less necessary consequence of defining "continuous" to mean "continuous at every real number in the domain". This is the right definition for functions on general metric spaces. I agree that it's probably confusing for people just starting calculus, which is why (I believe) most calculus textbooks define "continuous" to mean "continuous at every real number". But, it's a definition, so, at the end of the day, there's no right or wrong answer, so long as you then use the word consistently in accordance with the definition.

In terms of the example, $$x^2-9 \over x-3 $$ is continuous according to our definition; it's not continuous at 3, but it's not defined at 3 either.

I have no problem with changing to the more usual definition of continuous, anyway, if you want to solve the problem that way.--Warshall (talk) 15:05, 24 June 2009 (UTC)


 * I think we should change the definition. At the level of this book it is easier to define the unquantified term continuous to be continuous on the real line, since the style of the book so far is to not worry about specifying the domain of the functions that it defines, and when we want to be careful we can always say "continuous on ..." Thenub314 (talk) 19:40, 25 June 2009 (UTC)

Pencil lifting disconnect
Maybe I'm just failing to see it, but there seems to be a lack of logical connection between the intuitive notion of continuity meaning you can draw the graph of the function without lifting your pencil and the Intermediate Value Theorem (IVT). For example, consider the function $$f(x)=-x^2-1$$, and consider the points $$a=-1$$, $$b=1$$. The IVT tells us nothing whatsoever about the function in this case, because $$f(a)=0=f(b)$$, so there are no points between $$f(a)$$ and $$f(b)$$. I'm not saying that the IVT doesn't imply the intuitive notion of continuity, but the text doesn't seem to properly draw out the connection if there is one. Greenbreen (discuss • contribs) 15:40, 7 April 2011 (UTC)
 * After thinking about this further, I have removed the following text from the page,


 * Consider an even more degenerate example than the one I gave earlier: $$g(x)=0$$ with $$a=-1$$ and $$b=1$$.  Not only does the IVT not tell us anything using $$a$$ and $$b$$, but it will also tell us exactly squat for any other pair of points we choose in $$[a,b]$$.  Maybe someone could argue that the IVT requires continuity, and the pencil-lifting intuition can be derived in this case from the definition of continuity.  However, I think it is misleading to make the statement I quoted, so I removed it.  Greenbreen (discuss • contribs) 16:07, 7 April 2011 (UTC)