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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 18, 2013 11:43 PM
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On 9/28/2013 12:30 PM, Hetware wrote:
> I'm reading a 1953 edition of Thomas's Calculus and Analytic Geometry.
> In it he states that given:
> F(t) = (t^2-9)/(t-3)
>
> F(t) = (t-3)(t+3)/(t-3) = t+3 when t!=3.
>
> But F(t) is not defined at t=3 because it evaluates to 0/0.
>
> If someone were to ask me if (t^2-9)/(t-3) is defined when t=3, I would
> say it is because it can be simplified to t+3. Am I (and/or Thomas)
> engaging in meaningless hair-splitting regarding the question of F(3)
> being defined?


By using the parameter t, Thomas was baiting the physicist. He was
looking for a fight.

Let c[f(t)] be the assertion that for every real number t there is some
real number y such that f(t) = y, and also that for every real number t
limit[f(s), s->t] where s is a real number in the neighborhood of t. In
poor man's notation:

c[f(t)] => [y = f(t) y in R, all t in R] AND [limit[f(s), s->t] = f(t)]

'=>' is to be read 'implies'.

Let C[f(t)] be the statement that f(t) is continuous for all real numbers t.

C[f(t)] <=> [y = f(t) y in R, all t in R] AND [limit[f(s), s->t] = f(t)]

The equivalence relation '<=>' can be read as 'is defined as'.

Clearly c[f(t)] => C[f(t)].

There is nothing that I know of in mathematics which precludes the
assertion c[f(t)] if c[f(t)] does not lead to a contradiction. It is
usually accepted that 0/0 is indeterminate.

http://mathworld.wolfram.com/Indeterminate.html

Mathematicians are wont to equate /indeterminate/ with /undefined/. In
the case of a function such as f(x) = x/x, I challenge that equation.

0=1 is a contradiction, not an indeterminate form.

If we define f(x) as

f(x) = x/x and c[f(x)],

then limit[f(s), s->x] = f(x). Therefore, our definition tells us that
f(0) = limit[f(s), s->0] = 1.

Now, if we define f(x) = x/x, omitting c[f(x)], and then ask if C[f(x)]
is true, we are asking if

[y = f(x) y in R, all x in R] AND [limit[f(s), s->x] = f(x)].

In the process of determining whether f(x) is continuous we shall test
various propositions.

If we propose that C[f(t)=x/x] is not true, then we cannot use c[f(t)]
to find a value for f(0). That is to say, we find that f(0) = 0/0,
which is indeterminate. We have no other method of evaluation
applicable to f(0). We have no definition for f(0). We, therefore,
conclude that f(0) is undefined.

On the other hand, if we propose C[f(t)=x/x] we are assuming, for the
sake of argument, the assertion c[f(t)], and testing the results. In
this case, we find that /prima/ /facie/ f(0) = 0/0, but now we have the
additional rule limit[f(s), s->t] = f(t). Using that rule, f(0)=1.

I believe this is a profound observation, closely related to Russell's
Paradox.


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9/28/13
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Hetware
9/28/13
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