Epsilons and Deltas Book Error? Or My Error?

[Saladsamurai]

Homework Statement

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


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

$-c<|f(x)-L|<c$ 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:

$0<|f(x)-L|<c$

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

Why did they use the left-hand-side of the inequality if it was given that c > 0 ?

 

[snipez90]

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).

 

[Saladsamurai]

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

Thanks!

 

[Luongo]

what course is this

 

[Saladsamurai]

what course is this

This from a Calculus Textbook. Self Study.