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Proving the Squeeze when x->-infinity

  1. Oct 12, 2011 #1
    1. The problem statement, all variables and given/known data

    Prove the Squeeze theorem when x tends to -∞

    2. Relevant equations

    Squeeze theorem:
    f, g and h defined over (a-r , r+a) r>0 [itex]\forall[/itex] x[itex]\in[/itex] (a-r , r+a)
    f(x)[itex]\leq[/itex]g(x)[itex]\leq[/itex]h(x)
    if lim f(x) = lim h(x) = L as x[itex]\rightarrow[/itex]a
    then we have lim g(x) = L as x[itex]\rightarrow[/itex]a

    3. The attempt at a solution

    I am having a hard time finding the right interval.

    I am tempted to write:
    f, g and h defined over (-∞,+∞)
    Let ε>0, [itex]\exists[/itex] δ1, δ2>0:
    f(x)[itex]\leq[/itex]g(x)[itex]\leq[/itex]h(x)
    if lim f(x) = lim h(x) = L as x[itex]\rightarrow[/itex]-∞
    then we have lim g(x) = L as x[itex]\rightarrow[/itex]-∞

    but then if I follow the proof my prof gave me, I'd end up with
    [itex]\forall[/itex]x 0<|x- -∞|<δ1
    and 0<|x- -∞|<δ2

    Thanks in advance
     
  2. jcsd
  3. Oct 12, 2011 #2

    LCKurtz

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    You can't use a "δ" definition for x → ∞ or x → -∞. You need to start with a proper definition:

    [tex]\lim_{x\rightarrow -\infty}g(x) = L\hbox{ means given }\epsilon > 0
    \hbox{ you can find } N\hbox{ such that }|g(x) - L | < \epsilon\hbox{ for all } x < N[/tex]

    That is what you need to prove.
     
    Last edited: Oct 12, 2011
  4. Oct 17, 2011 #3
    Would the following be correct?

    f, g and h defined over I=(-∞,N) N ∈ ℝ
    If ∀ x ∈ I, we have:
    f(x)≤g(x)≤h(x) and lim f(x) = lim h(x) = L as x→-∞,
    then we have:
    lim g(x) = L as x→-∞

    lim f(x) = L as x→-∞ implies:
    [itex]\forall[/itex] ε>0, ∃ N1>0, ∀x, x<N1→ L-ε<g(x)<L+ε

    lim h(x) = L as x→-∞ implies:
    [itex]\forall[/itex] ε>0, ∃ N2>0, ∀x, x<N2→ L-ε<g(x)<L+ε

    Let N=max(N1,N2)

    [itex]\forall[/itex] ε>0, ∃ N>0, [itex]\forall[/itex] x<N → L-ε<f(x)≤g(x)≤h(x)<L+ε
    which proves
    [itex]\forall[/itex] ε>0, ∃ N>0, [itex]\forall[/itex] x<N → g(x)[itex]\in[/itex](L-ε , L+ε)

    therefore lim g(x)=L as x→-∞
     
  5. Oct 17, 2011 #4

    LCKurtz

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    The structure of your argument is good, but you are still not using the correct definition for a limit as x → -∞. Remember that, intuitively, f(x) and g(x) must get close to L when x becomes a "large negative number". You don't get that with your conditions on N I have highlighted. x less than some positive number doesn't qualify.

    Fix that with a couple of minor changes and you will be good to go.
     
  6. Oct 17, 2011 #5
    I think I've got it:

    lim f(x) = L as x→-∞ implies:
    ∀ ε>0, ∃ N1>0, ∀x, x<-N1→ L-ε<g(x)<L+ε

    lim h(x) = L as x→-∞ implies:
    ∀ ε>0, ∃ N2>0, ∀x, x<-N2→ L-ε<g(x)<L+ε

    Let N=max(N1,N2)

    ∀ ε>0, ∃ N>0, ∀ x<-N → L-ε<f(x)≤g(x)≤h(x)<L+ε
    which proves
    ∀ ε>0, ∃ N>0, ∀ x<-N → g(x)∈(L-ε , L+ε)

    therefore lim g(x)=L as x→-∞
     
  7. Oct 17, 2011 #6

    LCKurtz

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    OK, that's better. Or you could just say there is an N such that x < N implies... and take the minimum of N1 and N2. You don't need the N > 0 and the minus signs, although the way you have written it is not incorrect.
     
  8. Oct 17, 2011 #7
    OK. Thanks a lot for your help LCKurtz.
     
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