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
Then you would be wrong.
>because it can be simplified to t+3.
To get that result, you had to cancel the common factor t-3 in numerator and denominator. But that cancellation depends on the simplification
(t - 3)/(t - 3) = 1
which is only valid if t != 3.
In a first level algebra course (Elementary Algebra) where function concepts are not yet in play, the simplification
(t^2 - 9)/(t - 3)
= (t + 3)(t - 3))/(t - 3)
= t + 3
is allowed, without worrying about exceptional values of t for which the simplification fails.
But at the next level of algebra, algebraic expressions are often being regarded as functions, so more care is taken to identify those exceptional values.
>Am I (and/or Thomas) engaging in meaningless hair-splitting >regarding the question of F(3) being defined?
For functions, identifying the precise domain is key.
Thomas is correct.
Those hairs _need_ to be split.
All modern precalculus and calculus texts are careful (in the context of deciding whether functions are equal) to identify exceptional values where simplifications fail.