Register to reply

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

by Saladsamurai
Tags: book, deltas, epsilons, error
Share this thread:
Sep13-09, 12:21 AM
Saladsamurai's Avatar
P: 3,016
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 ?
Phys.Org News Partner Science news on
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100
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
Saladsamurai's Avatar
P: 3,016
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
Saladsamurai's Avatar
P: 3,016
Quote Quote by Luongo View Post
what course is this
This from a Calculus Textbook. Self Study.

Register to reply

Related Discussions
Epsilons and Deltas and Such Calculus & Beyond Homework 4
Measurement error analyses, fitting min/max slopes to data with error bars. Set Theory, Logic, Probability, Statistics 2
Error in book? Please confirm. General Math 4
Error in book ? General Math 12
Proof using epsilons and deltas. Calculus 5