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Limit proof help

  1. Jul 7, 2008 #1
    1. The problem statement, all variables and given/known data

    Let f and g be functions such that 0≤f(x)≤g(x) for every x near c, except possibility at c. Use the definition of a limit to prove that if lim(x→c)g(x)=0 then lim (x→c)f(x)=0.

    This is my attempt at this proof and I'm very poor at it. Can someone check and give me some pointers? Thanks!

    2. The attempt at a solution
    Definition: Limit lim(x→a)f(x)=L if and only if, given ϵ>0, there exists δ>0 such that 0<|x-a|<δ implies that |f(x)-L|<ϵ.
    Given the definition of a limit we can state that lim(x→c)g(x)=0 has the following properties:
    0<|x-c|<δ_0 Implies that |g(x)-0|<ϵ_0.
    Thus lim(x→c)f(x)=0, 0<|x-c|<δ_1 implies that |f(x)-0|<ϵ_1.
    By the original given statement that 0≤f(x)≤g(x), we can argue that ϵ_1< ϵ_0 and δ_1<δ_0
    Also by this given since lim(x→c)g(x)=0 and 0≤f(x)≤g(x)
    The value of f(x) is between 0≤g(x)
    Therefore |g(x)-0|<ϵ_0= |g(x) |<ϵ_0. Thus δ_0= ϵ_0.
    Since 0<|x-c|<δ_0 It follows that |g(x)-0|= |g(x) |<δ_0 where δ_0= ϵ_0. Therefore the limit lim(x→c)g(x)=0 holds true. Since this is true for lim(x→c)g(x)=0 it follows that it must also be true for lim(x→c)f(x)=0 because of the given 0≤f(x)≤g(x) properties of the given function and since ϵ_1< ϵ_0 and δ_1<δ_0.
  2. jcsd
  3. Jul 7, 2008 #2
    i don't believe your proof is correct

    "Therefore |g(x)-0|<ϵ_0= |g(x) |<ϵ_0. Thus δ_0= ϵ_0."

    i have no idea why this is necessary.

    if you're already on limits of functions, it must mean you past limits of sequences.
    The way i did this problem follows very similarly to the proof of the squeeze theorem for sequences.

    if you have done that, i suggest you analyze it.

    i don't think your proof is started correctly. i'll get you on the right track, but i won't finish it for you.

    so we know:
    1) 0 less than equal to f(x) less than equal to g(x)
    2) limit x->c g(x) = 0

    to prove by definition, WMST for all epsilon > 0, there exists a delta > 0, such that for all real numbers x, IF 0< |x-c| < delta, then |f(x)-0|< epsilon

    let epsilon be an arbitrary real number > 0... we must find this delta > 0 [hence this is an existence proof, where you must find delta and then verify]

    let's start from the end. we want |f(x)-0|< epsilon
    well, |f(x)-0|< epsilon if and only if - epsilon < f(x)-0 < epsilon

    so we need to satisfy both sides of the inequality - epsilon < f(x)-0 < epsilon

    The left side is very simple...
    from our assumption that 0 less than equal f(x) less than equal g(x)
    and our other assumption that epsilon > 0...
    we can claim that -epsilon < 0 (multiplying the above line by -1 on both sides)
    therefore -epsilon < 0 < f(x) (combining the above lines)
    hence -epsilon < f(x) as required

    The right side...

    note that since we have an arbitrary epsilon > 0, and also from our assumption that lim g(x) = 0,

    we know that there exists a different delta, which i'll denote as delta1. such that for all reals x, IF 0< |x-c| < delta1, then |g(x)-0|< epsilon

    this is where i will stop. i won't fill in the details, but i will give you the simple answer
    Let the original delta we were looking for, be this new delta1
    That is... let delta=delta1

    continue by working on the the end of IF 0< |x-c| < delta1, then |g(x)-0|< epsilon

    expand that "end" similar to how we expanded it earlier.

    you should end up with -epsilon < f(x) < epsilon
    which is what we wanted...

    but note that all of this work (for the right hand side) was based on the assumption that IF 0< |x-c| < delta1

    since this proof is an existance "column" proof where you have to find the original delta, then verify the proposition.
    when you verify the proposition "for all real numbers x, IF 0< |x-c| < delta, then |f(x)-0|< epsilon"
    you're basically doing a "conditional" proof...
    i'm sure you're aware, to do that you assume the hypothesis... and must verify the conclusion
    when you do assume the hypothesis... and you have that delta=delta1 ... it gives you the the condition needed to make the assumption in the line before this paragraph...

    hope that clears it up
    Last edited: Jul 7, 2008
  4. Jul 7, 2008 #3
    Thanks for the help, but unfortunately my teacher didnt really go through this very well nor touched on limit sequences.

    "Therefore |g(x)-0|<ϵ_0= |g(x) |<ϵ_0. Thus δ_0= ϵ_0."

    my thoughts on this was that since g(x) is actually bigger then f(x) we must take the condition to do the limit proof by definition on this function first before anything could be assumed for f(x). I was merely trying to distinguish that these epsilons were different then the ones being looked at for the two separate functions. I'm still unclear why thats not necessary. Arent they separate?

    Thanks again
  5. Jul 7, 2008 #4


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    Homework Helper

    This is really a whole lot easier than that.

    I like your start: from the definition of a limit you know that

    [tex]\lim_{x\to c}g(x)=0\Leftrightarrow\forall\epsilon >0\,\exists\delta>0\mbox{ such that }0<|x-c|<\delta\Rightarrow |g(x)-0|<\epsilon[/tex]

    Now insist that [tex]0<|x-c|<\delta[/tex] means "x is near c" so that after we insert [tex]0\le f(x)\le g(x)[/tex] into the above we have

    [tex]\forall\epsilon >0\,\exists\delta>0\mbox{ such that }0<|x-c|<\delta\Rightarrow [f(x)-0|\le |g(x)-0|<\epsilon \Leftrightarrow \lim_{x\to c} f(x) =0[/tex]

    (A note on notation:
    [tex]\forall[/tex] means "for all" or "for every"
    [tex]\exists[/tex] means "there exists"
    [tex]\Rightarrow[/tex] means "implies that"
    [tex]\Leftrightarrow[/tex] means "if, and only if" (iff for short).)
    Last edited: Jul 7, 2008
  6. Jul 7, 2008 #5
    well, f(x) and g(x) are indeed two different functions...
    but what you're trying to prove does not require using "two different epsilons"
    the problem only states to prove that the limit of f(x) as x approaches c = 0
    write out the symbolic form of that...
    which is, for all epsilon greater than 0, there exists a delta greater than 0, for all real numbers x, etc...

    the first step of this proof is to assume that there is an arbitrary epsilon greater than 0

    Now with this hypothesis, this single arbitrary epsilon that is greater than 0, can be used in the definition of the limit of g(x) as x approaches c.
    we don't have to prove that the limit of g(x) as x approaches c = 0... it is already stated that this is the case... (that is... you probably should USE, not prove, the fact that lim g(x) = 0)

    if you write out the definition of the limit of g(x) = 0... and NOTE that we HAVE AN EPSILON > 0... apply the definition of the limit, you get results

    which is used in my original post
  7. Jul 7, 2008 #6
    Ah I see now. I think I got it now. Its definitely a lot easier then i set it out to be. Didnt notice the IF-THEN and givens. Many thanks for all the help its been greatly appreciated.
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