Hetware wrote: > On 10/6/2013 11:45 AM, Peter Percival wrote: >> Hetware wrote: >> >>> OK. I turned this around and can see the problem. I was assuming f(t)= >>> (t^2-9)/(t-3) is continuous, which is not justified from the context. At >>> t=3, x can be anything in t^2 - 9 = x (t - 3). >> >> If f is a function with domain the most inclusive set of reals possible, >> then f is continuous. >> > > If I say f(t) is continuous for all real numbers, and define > f(t)=(t^2-9)/(t-3) for all t!=3,
That's back to front. You define f thus and then deduce that f is continuous.
> then limit[f(t),t->3]=6. So f(3)=6, by > the definition of continuity.
No, f(3) doesn't equal anything, you wrote f(t)=(t^2-9)/(t-3) for all t!=3.
> A function thus defined is identical to > g(t)=t+3.
You could say, more directly, g:R->R is defined by
g(t) = f(t) if t=/=3 = 6 otherwise,
and then prove that g is continuous.
> The assertion that the f(t) is continuous determines the value of f(t) > at t=3 without the need for further elaboration. So, I concede that I > was wrong in believing that f(t) defined as f(t)=(t^2-9)/(t-3) is > identical with g(t)=t+3. The continuous function defined as > f(t)=(t^2-9)/(t-3) everywhere the rhs is defined is identical with > g(t)=t+3.
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