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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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Richard Tobin

Posts: 1,267
Registered: 12/6/04
Re: Is (t^2-9)/(t-3) defined at t=3?
Posted: Oct 20, 2013 8:43 AM
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In article <l40bu1$sqi$1@news.albasani.net>,
Peter Percival <peterxpercival@hotmail.com> wrote:

>> Let f(x) be continuous over R, and f(x) = x/x for all x in R where x/x
>> is a determinate form?


That's a perfectly good definition, except that you have to prove
that such a function exists (which is easy).

>So the domain of f is R\{0} and f(x) = 1. Since f isn't defined at 0 it
>cannot be continuous there.


There is a unique function on R that is continuous and whose value
is x/x where that is defined. So that can be used as a definition.

In this case, you could call f(x) the "continuous continuation" of
x/x, on the analogy of "analytic continuation".

>It makes no sense to say let f be continuous, and _then_ define it.
>Define f first and then prove it to be continuous if it is so.


No, there's no problem with saying that a function is continuous as
part of its definition. You just (as always) have to show that the
definition is consistent.

-- Richard


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9/28/13
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Ronald Benedik
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Peter Percival
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Virgil
10/18/13
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10/19/13
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Peter Percival
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Hetware
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10/19/13
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