
#1
Sep1309, 12:21 AM

P: 3,012

1. The problem statement, all variables and given/known data
Given f(x) = mx + b, m > 0, L = (m/2) + b, x_{o} = 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  x_{o} < [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 righthandside 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: [itex]0<f(x)L<c[/itex] 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 lefthandside of the inequality if it was given that c > 0 ? 



#2
Sep1309, 12:47 AM

P: 1,106

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). 



#3
Sep1309, 12:51 AM

P: 3,012

Oh yeah! We define epsilon to be a positive number.
That wAs stupid of me! Thanks! 



#4
Sep1309, 01:13 AM

P: 120

Epsilons and Deltas:: Book Error? Or My Error?
what course is this



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