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!

Homework Help: Basic limit proof of limit equivalence

  1. Oct 22, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove that limx[itex]\rightarrow[/itex]cf(x)=L if and only if limh[itex]\rightarrow[/itex]0f(x+h)=L.

    2. Relevant equations

    3. The attempt at a solution

    I think this is a simple problem, but I am getting caught up in the middle, as I'm not sure if my procedure is a valid way to prove the statement.

    Suppose limx[itex]\rightarrow[/itex]cf(x)=L.

    Then for each [itex]\epsilon[/itex]>0 there is a [itex]\delta[/itex]>0 so that

    if |x-c|<[itex]\delta[/itex] then |f(x)-L|<[itex]\epsilon[/itex].

    Let x=c+h. Then if |x-c|<δ [itex]\Rightarrow[/itex] |c+h-c|<δ [itex]\Rightarrow[/itex] |h|<δ. Furthermore, |f(c+h)-L|<ε. And since we assumed limx[itex]\rightarrow[/itex]cf(x)=L, it follows then that limh[itex]\rightarrow[/itex]0f(c+h)=L.*

    So my questions is this: it is valid to simply let x=c+h as I did?

    *Note, I did not prove the other case, but I just wrote this out so you guys can get a good idea of what my argument is and tell me if it is wrong or not.
  2. jcsd
  3. Oct 22, 2013 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    This is the right approach, but your arrows don't point the way you want them to. To show that if h goes to zero, f(c+h) goes to L, you need to start with: if |h| < δ, things happen. You started by assuming that |x-c| < δ, then showed for your choice of defining h that |h| < δ as well, which isn't good enough. But all the stuff you wrote down is reversible, so you should be able to start with |h| < δ and then use what you know about f(x) when x=c+h.
  4. Oct 22, 2013 #3
    Thank you very much Office_Shredder! One question though: I thought that I had to prove it in both orders, since it is an if and only if statement. Your route certainly seems more sensible but if I were to write the complete solution, wouldn't I assume in one case that as h goes to zero f(c+h) goes to f(c) and show that f(x) going to f(c) follows, and then start over and assume that as x goes to c f(x) goes to f(c) and show that f(c+h) going to f(c) follows from there? What I am asking is if I wanted to prove A iff B, wouldn't I do:

    If A is true, then B follows.
    If B is true, then A follows.

    Thus, A is true iff B is true. ? (Hopefully that example makes more sense than the above).

    Again, thank you!
  5. Oct 22, 2013 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yes, but it's important to prove if A then B in the section where you are claiming if A then B. If you want to assume
    [tex] \lim_{x\to c} f(x) = L [/tex]
    and then prove that
    [tex] \lim_{h\to 0} f(c+h) = L [/tex],
    then you better at some point have h be an arbitrary number such that |h| < δ
  6. Oct 22, 2013 #5
    Ohhhhh! I see. Thank you!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted