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Topic: Is (t^2-9)/(t-3) defined at t=3?
Replies: 166   Last Post: Oct 30, 2013 9:41 AM

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 magidin@math.berkeley.edu Posts: 11,749 Registered: 12/4/04
Re: Is (t^2-9)/(t-3) defined at t=3?
Posted: Oct 8, 2013 6:25 PM

On Tuesday, October 8, 2013 3:13:56 PM UTC-5, quasi wrote:
> Arturo Magidin wrote:
>

> >
>
> >A real valued function of real variable f(t) is
>
> >"continuous at x=a" if and only if:
>
> >
>
> >(i) It is defined at a;
>
> >(ii) The limit of f(t) as t approaches a exists; and
>
> >(iii) the value of the function at a equals the value of
>
> >the limit of f(t) as t approaches a.
>
> >
>
> >A real valued function of real variable f(t) is "continuous
>
> >at all real numbers" or "continuous everywhere" if and only if
>
> >it is continuous at x=a for all real numbers a.
>
>
>
> In the Precalculus/Calculus context, the above is consistent
>
> with the definitions I'm familiar with.
>
>
>
> On the other hand, I've never seen the usage you describe
>
> below, at least not in the current context.

Part of the issue here is what to do with isolated points of the domain, or more generally points in which there is no neighborhood of the point contained in the domain.

In Precalculus/Calculus, these kinds of things do not usually occur, except in the special case of endpoints of intervals, because we seldom consider functions whose domains have isolated points.

The usual definition of "continuous at x=a" using epsilon and deltas that I am familiar with is:

A function f is continuous at x=a if and only if f is defined at a, and for all e>0 there exists d>0 such that for all x **in the domain of f**, if |x-a|<d, then |f(x)-f(a)|<e.

This allows you to talk about continuity at isolated points, and continuity of restrictions to arbitrary subsets of the domain.

I am aware that the distinction between "continuous on X" and "continuous" is not generally discussed in many precalculus textbooks (they have enough trouble as it is).

>
> >A real valued function of real variable f(t) is "continuous"
>
> >if and only if it is continuous at all points in its domain.
>
>
>
> In the Precalculus/Calculus context, here is my take ...
>
>
>
> A function f is continuous on an interval if it is continuous
>
> at all points of that interval (with one-sided continuity
>
> acceptable if the interval has an endpoint).

Yes, I agree with that.

> A function f is "continuous" if the domain of f is an interval
>
> and f is continuous on that interval.

But not with that. That would mean that, for example,

f(x) = sqrt(1-x^2)

is a fortiori not continuous, since its domain is (-oo,-1] \/ [1,oo), which is not an interval.

>
>
>
> Examples:
>
>
>
> The function x^2 is continuous since its domain is the
>
> interval (-oo,oo) and it's continuous on (-oo,oo).
>
>
>
> The function sqrt(x) is continuous since its domain is the
>
> interval [0,oo) and it's continuous on [0,oo).
>
>
>
> The function ln(x) is continuous since its domain is the
>
> interval (0,oo) and it's continuous on (0,oo).
>
>
>
> The function 1/x is _not_ continuous since its domain is not
>
> an interval. Of course one _can_ say that 1/x is continuous
>
> on the each of the intervals (-oo,0) and (0,oo).
>
>
>
> The function tan(x) is _not_ continuous since its domain is
>
> not an interval. Of course one can say that tan(x) is
>
> continuous on the interval (-Pi/2,Pi/2) and more generally
>
> tan(x) is continuous on any interval of the form
>
> (a - Pi/2, a + Pi/2) where a = k*Pi for some integer k.

I consider tan(x) to be a continuous function, because it is continuous wherever it is defined. This does *not* mean that tan(x) is continuous at each number.

So, for example, under the usage I mention, a sum, difference, product, and quotient of continuous functions is necessarily continuous, whatever the domains may be; in particular, a rational function is "continuous", though it need not be "continuous everywhere". In essence, whether the function is "continuous" is an intrinsic property of the function; whether a function is "continuous on X" is a *contextual* property of the function in relation to X.

This is signaled by the fact that "continuous" refers only to the function, whereas "continuous on X" specifies both the function and the context.

> By your usage, all of the above functions, including the
>
> functions 1/x and tan(x) would be regarded as "continuous",
>
> without qualification.

Yes.

> I've never seen that usage.
>
>
>
> Worse, it jars with the naive, Precalculus concept of
>
> continuity, typically expressed as:
>
>
>
> "A function is continuous if the graph has no breaks."

Ah, but that naive notion jars with the idea that you can have functions that are continuous at exactly one point and nowhere else, as well (e.g., xsin(1/x) for x=/=0, and 0 for x=0), and yet such functions exist. Jarring with naive notions is not in and of itself a bad thing, given that naive notions are so often wrong.

> or equivalently,
>
>
>
> "One can trace the graph of the function without lifting
>
> the pen from the paper."

See above. The function f(x) = xsin(1/x) for x=/=0 and f(0)=0 is undeniably continuous at 0; but that idea simply cannot be reconciled with the naive notion.

(In fact, Spivak brings up this example to note that the naive notion of "a continuous function is one whose graph can be drawn without lifting the pencil from the paper" is "a little too optimistic")

>
>
>

> >So the function f(t) = (t^2-9)/(t-3) is continuous;
>
>
>
> Very strange.
>
>
>
> Can you find a reference to a textbook at the Precalculus
>
> or Calculus level which has a definition by which a function
>
> similar to the function f above would be regarded as a
>
> continuous function?

Don't know about Precalculus (I would be hard pressed to find a precalculus textbook that even discusses limits appropriately, let alone the subtle notions of continuity). It matches the use in Bourbaki (Elements of Mathematics, General Topology, Chapter 1, Section 2, part 1) (Yes, I know Bourbaki is not precalculus or calculus)

A mapping f from a topological space X into a topological space X' is said to be continuous at x_0 in X if, given any neighborhood V' of f(x_0), there exists a neighborhood V of x_0 in X such that the relation x in V implies f(x) in V'.

A mapping of a topological space X into a topological space X' is said to be continuous if it is continuous at each point of X.

The notions are extended to real-valued functions in IV.5.1.

Won't help you because it is in Spanish, but this is also the convention used in Arizmendi, Carrillo, and Lara's "Calculo Diferencial e Integral" (Editorial Trillas), which was the textbook used when I took the course as an undergraduate.

While I don't find other books on my shelf with the explicit usage I give, I also find no books on my shelf with the usage *you* give on "continuous" to mean "defined on an interval and continuous on that interval". Spivak always says "continuous on..."; Hughes-Hallet, Gleason, and McCallum suddenly use "continuous" on page 57 of their Single Variable 5th Edition without every having said what it means to be "continuous"(they have only talked about "continuous at x=c" and "continuous on an interval"); Anton, Bivens, and Davis only use "continuous on...", except for one informal statement that they specify and highlight as being "informal". Stewart talks about "continuous on..." and "continuous at...", never about just "continuous".

> f(t) = (t^2 - 9)/(t - 3)
>
>
>
> would be called a continuous function.
>
>
>
> I don't think so.

I have never encountered your definition of "f is continuous" (without specification of where) that implies the domain is an interval, either in a precalculus, calculus, or more advanced level, that I can remember. I would expect, in fact, that most books would consider sqrt(1-x^2) to be "a continuous funcion", despite the domain being disconnected.

I've encountered mine, though I free admit that it is not widespread at that level; most precalculus and calculus textbooks never refer to a function as just being "continuous" (unless they are being purposely careless, like Anton/Bivens; or just plain sloppy, like Hughes-Hallet/Gleason/McCallum, never saying what they mean).

--
Arturo Magidin

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