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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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Peter Percival

Posts: 1,301
Registered: 10/25/10
Re: Is (t^2-9)/(t-3) defined at t=3?
Posted: Oct 9, 2013 4:44 AM
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Hetware wrote:

>>
>> I.e. for t!=3, so f isn't actually defined for all t, and even less it
>> is continuous for all t.
>>

>
> Wrong. f(t) is defined at t=3.


Let's try it: f(3) = (3^2 - 9)/(3 - 3) = 0/0. Undefined like I said.

> The assumption of continuity requires

There is no "assumption of continuity". f is, indeed, continuous at
every point in its domain, but that is to be proved not just assumed.

limit[f(t), t->c] can exist even though f(c) doesn't. Your f with c = 3
demonstrates that.

> f(c) = limit[f(t),t->c].
>
> For example, the position of a particle along the x axis is given by
> f(t) = (t^2-9)/(t-3), where t represents time.


A particle is what? Time is what? Those are matters for physicists not
mathematicians. Is a particle a geometrical point? Is an instant a
real number?


--
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9/28/13
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Hetware
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Peter Percival
9/29/13
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quasi
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Hetware
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Richard Tobin
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Hetware
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Hetware
10/6/13
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Hetware
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10/13/13
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10/19/13
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10/20/13
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Peter Percival
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Virgil
10/18/13
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Hetware
10/19/13
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Peter Percival
10/19/13
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Peter Percival
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Peter Percival
10/19/13
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Hetware
10/19/13
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10/19/13
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magidin@math.berkeley.edu
10/19/13
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Hetware
10/19/13
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10/20/13
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