The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Epsilon - Delta Proofs

Date: 1/23/96 at 13:47:58
From: Catherine Miller
Subject: Def. of Limit?

lim f(x) = L
0<|x-c|< d (delta), then |f(x)-L|< E (epsilon)


  lim(2x - 5) = 1

find delta such that |(2x - 5) - 1| < 0.01  whenever 0 < |x - 3| < delta
                the f(x) /       \the L                  ?/    \the c

I'm confused about the f(x) in the epsilon function and the x in 
the delta function. Does the delta (x) become a value when solving 
these equations?

Date: 6/20/96 at 9:56:40
From: Doctor Jerry
Subject: Re: Def. of Limit?

To show that the the limit of 2x-5 is 1 as x approaches 3, you 
must show that for each choice of E you can find a delta so that
|(2x - 5) - 1| < E whenever 0 < |x - 3| < delta.  

With your choice of E=0.01, let's find a suitable delta by doing 
some scratch work first. We want |(2x - 5) - 1| < 0.01.  Rewrite 
this as |2x - 6| < 0.01 and, again, as  |x - 3| < 0.01/2.  

In this form, it appears that if we take delta =0.01/2, then 
|(2x - 5) - 1| < 0.01.

In fact, if you replace 0.01 by E, you can in the same way, obtain 
a general procedure for choosing delta, once given an E.  Namely, 
choose delta=E/2.

I hope this answers your question.  If not, write back to us.

-Doctor Jerry,  The Math Forum

Associated Topics:
High School Analysis

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.