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Epsilons and Deltas:: Book Error? Or My Error?

by Saladsamurai
Tags: book, deltas, epsilons, error
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Sep13-09, 12:21 AM
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1. The problem statement, all variables and given/known data

Given f(x) = mx + b, m > 0, L = (m/2) + b, xo = 1/2 , [itex]\epsilon = c >0[/itex] find (a) an open interval on which the inequality
|f(x) - L| < [itex]\epsilon[/itex]
holds. Then find (b) [itex]\delta[/itex] such that 0 < |x - xo| < [itex]\delta\Rightarrow[/itex] |f(x) - L| < [itex]\epsilon[/itex]

Here is my problem with the book's solution. Since the condition [itex]\epsilon=c>0[/itex] was given, I only used the right-hand-side of the inequality:

[itex]-c<|f(x)-L|<c[/itex] because to me it did not make sense to solve the inequality under a condition that cannot be. Instead, I chose to write the above inequality as:


But the text gave answer of (a) [itex](\frac{1}{2}-\frac{c}{m}, \frac{c}{m}+\frac{1}{2})[/itex] and (b) [itex]\delta = c/m[/itex]

Why did they use the left-hand-side of the inequality if it was given that c > 0 ?
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Sep13-09, 12:47 AM
P: 1,104
The book is right, but I don't know why it introduces the unneeded term c. I'm not sure if they are trying to trick you, but epsilon, or c for that matter, is just an arbitrary positive number. Look at the inequality |f(x) - L| < e again (e is just an arbtirary positive number; we usually think of it as very small). You interpreted this correctly in the previous limit question I helped you out with. All this inequality is saying is that f(x) is within a distance e from L.

Remember, |f(x) - L| < e is equivalent to L - e < f(x) < L + e. If you choose to write it in the latter form, you have to drop the absolute value signs (which I think is what tripped you up).
Sep13-09, 12:51 AM
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Oh yeah! We define epsilon to be a positive number.
That wAs stupid of me!


Sep13-09, 01:13 AM
P: 120
Epsilons and Deltas:: Book Error? Or My Error?

what course is this
Sep13-09, 10:08 AM
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P: 3,016
Quote Quote by Luongo View Post
what course is this
This from a Calculus Textbook. Self Study.

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