- #1

- 11

- 0

I'm looking for a good, general explanation of how to do delta/epsilon proofs. I've searched all over the web but this stuff is just really confusing. Let me start with a problem, show my attempt at a solution, and then maybe you guys can explain it in a way that will make sense to me.

Prove that: [tex]\lim_{x \rightarrow 2} 3x + 1 = 7 [/tex]

If [tex](0 < |x-a| < \delta)[/tex] implies [tex](|f(x) - L| < \epsilon)[/tex], then [tex]\lim_{x\rightarrow a} f(x) = L [/tex].

We want to prove that: [tex] \lim_{x \rightarrow 2} 3x + 1 = 7 [/tex] To do this, we must show that [tex](0 < |x- 2| < \delta)[/tex] implies: [tex](|3x + 1 - 7| < \epsilon)[/tex] So I start by taking: [tex]3x + 1 = 7[/tex] and simplifying it to [tex]3x - 6 = 0[/tex] Which we can set to being less than epsilon as a way of choosing a delta, thus [tex]|3x - 6| < \epsilon [/tex] which means that [tex]3|x-2| < \epsilon[/tex] so I can choose [tex]\delta = \frac{\epsilon}{3}[/tex]. Now I can assume that [tex]0 < |x - 2| < \delta[/tex], and since [tex] \delta = \frac{\epsilon}{3}[/tex] this means that [tex] 0 < |x - 2| < \frac{\epsilon}{3}[/tex]

At this point, I just draw a blank. I feel as if I have everything I need to complete the proof, but I'm missing something, and all the explanations I read don't seem to help. Can someone please help me get this through my thick skull?

## Homework Statement

Prove that: [tex]\lim_{x \rightarrow 2} 3x + 1 = 7 [/tex]

## Homework Equations

If [tex](0 < |x-a| < \delta)[/tex] implies [tex](|f(x) - L| < \epsilon)[/tex], then [tex]\lim_{x\rightarrow a} f(x) = L [/tex].

## The Attempt at a Solution

We want to prove that: [tex] \lim_{x \rightarrow 2} 3x + 1 = 7 [/tex] To do this, we must show that [tex](0 < |x- 2| < \delta)[/tex] implies: [tex](|3x + 1 - 7| < \epsilon)[/tex] So I start by taking: [tex]3x + 1 = 7[/tex] and simplifying it to [tex]3x - 6 = 0[/tex] Which we can set to being less than epsilon as a way of choosing a delta, thus [tex]|3x - 6| < \epsilon [/tex] which means that [tex]3|x-2| < \epsilon[/tex] so I can choose [tex]\delta = \frac{\epsilon}{3}[/tex]. Now I can assume that [tex]0 < |x - 2| < \delta[/tex], and since [tex] \delta = \frac{\epsilon}{3}[/tex] this means that [tex] 0 < |x - 2| < \frac{\epsilon}{3}[/tex]

At this point, I just draw a blank. I feel as if I have everything I need to complete the proof, but I'm missing something, and all the explanations I read don't seem to help. Can someone please help me get this through my thick skull?

Last edited: