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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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Hetware

Posts: 148
Registered: 4/13/13
Re: Is (t^2-9)/(t-3) defined at t=3?
Posted: Oct 14, 2013 3:23 PM
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On 10/14/2013 4:01 AM, Peter Percival wrote:
> Hetware wrote:
>

>>
>> The statement "Let f(t) be a continuous function for all real numbers t"
>> has a concise meaning.

>
> There's nothing wrong with that (one can easily prove that there are
> such functions). What is verboten is defining f with some expression in
> t and _then_ assuming it is continuous.
>


The more I think about this, the more ambiguous it seems.

One way to prove something in mathematics is to propose the opposite,
and show that the proposition leads to a contradiction. So let's try
that with the definition f(t) = (t^2-9)/(t-3) without giving continuity
as part of the definition.

Now, for the purpose of finding a contradiction, let us propose that
f(t) is globally continuous. What does that entail? f(t) has a
definite finite value for every real number t, and that value is f(t) =
limit[f(p), p->t].

Now, I concede that (t^2-9)/(t-3) = 0/0 when t = 3, and 0/0 is
meaningless. It is not, however, a contradiction. Furthermore, the
proposition of continuity gives us an alternative means at evaluating f(3).

So, one point of view is that f(t) is undefined at t = 3 because the
rule given for f(t) provides no value for f(3).

The other point of view is to say: bah! I say it's continuous, and as
soon as I say it's continuous, it satisfies the definition of continuity.

So we begin with two competing propositions.

Given f(t) defined by f(t) = (t^2-9)/(t-3) for the most inclusive domain
of real numbers t:

A) f(t) does not have a definite finite value such that f(t) =
limit[f(p), p->t] for all real numbers t.

B) f(t) has a definite finite value such that f(t) = limit[f(p), p->t]
for all real numbers t.

In support of A) we have (t^2-9)/(t-3) = 0/0 when t = 3. The definition
fails to provide a value for t = 3. In this case we have tacitly
assumed "discontinuous until proven continuous".

In support of B) we begin with the proposition that f(t) is globally
continuous, and seek a contradiction. IOW "continuous until proven
discontinuous". Sure, if we simply plug t = 3 into the expression
(t^2-9)/(t-3), we get f(3) = 0/0. But we can use the rule (t^2-9)/(t-3)
in conjunction with the assumption of continuity to evaluate f(3) by
another means. So we may contend that f(t) = (t^2-9)/(t-3) does,
indeed, provide a rule for determining a definite finite value such that
f(t) = limit[f(p), p->t] for all real numbers t.

It seems that in the case A) we are treating the set of real numbers as
isolated entities, more or less derived from the integers through a
number of extensions to the basic rules of combination. As such, we do
not consider any neighborhood of a point in the domain when we evaluate
f(t).

In case B) we are treating t as a position along a continuum, and bring
its neighborhood into consideration. To my way of thinking, both
approaches have merit.

There may be a compelling argument in favor of A), but "go back to high
school" doesn't work for me. I dropped out with a 0.636/4.0 GPA, and
fragments of a 10th grade education.

Thomas introduces the formal definition of continuity at the end of the
chapter subsequent to the one defining f(t). So, by the order of his
development f(t) = (t^2-9)/(t-3) was defined prior to the the definition
of continuity.

On the other hand, the definition of continuity is not logically
dependent on f(t) = (t^2-9)/(t-3), so applying it to this case doesn't
seem to pose circular reasoning. It is still unclear to me that
(t^2-9)/(t-3) = 0/0 when t = 3 proves f(t) = (t^2-9)/(t-3) is
discontinuous. I understand that is generally accepted to be the case,
especially in introductory calculus. But I haven't been able to come up
with any clear logical development that convinces me that it is
necessarily the case.




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9/28/13
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Hetware
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