- #1
Buri
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Is Michael Spivak Wrong!?
My text says,
In general, if ε > 0, to ensure that
|x²sin(1/x)| < ε,
we need only require that
|x| < ε and x ≠ 0, provided that ε ≤ 1. If we are given an ε which is greater than 1 (it might be, even thought it is "small" ε's which are of interest), then it does not suffice to require that |x| < ε, but it certainly suffices to require that |x| < 1 and x ≠ 0.
I don't see why if ε > 1 then it doesn't work. I've tried finding a counter example, but it always seem to work. Can someone please help me out?
EDIT:
x²sin(1/x) ≈ x when x → ∞ so it would seem that when x isn't very large then x²sin(1/x) will always be less then x and hence less then ε, even if ε were greater than 1.
ANY HELP?? Please?
BTW, this is Calculus by Spivak page 92-93.
My text says,
In general, if ε > 0, to ensure that
|x²sin(1/x)| < ε,
we need only require that
|x| < ε and x ≠ 0, provided that ε ≤ 1. If we are given an ε which is greater than 1 (it might be, even thought it is "small" ε's which are of interest), then it does not suffice to require that |x| < ε, but it certainly suffices to require that |x| < 1 and x ≠ 0.
I don't see why if ε > 1 then it doesn't work. I've tried finding a counter example, but it always seem to work. Can someone please help me out?
EDIT:
x²sin(1/x) ≈ x when x → ∞ so it would seem that when x isn't very large then x²sin(1/x) will always be less then x and hence less then ε, even if ε were greater than 1.
ANY HELP?? Please?
BTW, this is Calculus by Spivak page 92-93.