1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Limits - Apostol book

  1. Mar 18, 2013 #1
    Hello guys, I am stuck in page 129 on Calculus Vol I - Apostol book. I would like to know if there is anybody here who can help me. I am not a mathematician, so It might be a simple transformation but I am not going through it.

    He states that:

    lim(x->p) f(x) = A is equivalent to say that:

    lim(x->p) ( f(x) - A) = 0 OR

    lim(x->p) | f(x) - A| = 0

    I can not make these transformations algebraically, how can it be done?

    Thanks for any help
    Cheers
     
  2. jcsd
  3. Mar 18, 2013 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Would you agree a=b, a-b=0 and |a-b|=0 all mean the same thing? That's really almost the whole story. There is nothing complicated going on here.
     
    Last edited: Mar 18, 2013
  4. Mar 18, 2013 #3
    The book's definition of a limit is such that:

    | f(x) -A | < ε whenever 0 < | x - p | < δ

    So it presents the three equations (on the previous post) and it says: The equivalence becomes apparent as soon as we write each of these statements in the ε, δ terminology.

    That is the core of my doubts. How to write those three equations.

    Sorry that I wasn't clear enough in the first post.
     
  5. Mar 18, 2013 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    The definition of lim(x->p) f(x)=A is for all ε>0 there exists a δ>0 such that |f(x)-A|<ε whenever 0<|x-p|< δ. Don't leave out the quantifiers. Now follow their advice and write out what lim(x->p) (f(x)-A)=0 means.
     
  6. Mar 18, 2013 #5
    that is exactly the point where I am confused.

    Since we have:
    lim(x->p) f(x) = A, then, in my opinion, we would have

    lim(x->p) ( f(x) - A) = 0, since A is defined in the line above:

    lim(x->p) ( f(x) - lim(x->p) f(x) ) = 0

    And I don't know how to go any further from here. ;/

    Alternatively, iff you consider:
    lim(x->p) ( f(x) -A ) = 0, as we know |f(x) -A| < ε
    lim(x->p) ( ε ) would be ε, not zero.
     
    Last edited: Mar 19, 2013
  7. Mar 19, 2013 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    If lim(x->p) g(x)=0 then what's the definition of that? Spell it out for me. Then put g(x)=f(x)-A.
     
  8. Mar 19, 2013 #7
    Thank you, that was good.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted