Proving the Squeeze when x->-infinity

[frenchkiki]

Homework Statement

Prove the Squeeze theorem when x tends to -∞

Homework Equations

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

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, $\exists$ δ₁, δ₂>0:
f(x)$\leq$g(x)$\leq$h(x)
if lim f(x) = lim h(x) = L as x$\rightarrow$-∞
then we have lim g(x) = L as x$\rightarrow$-∞

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

Thanks in advance

 

[LCKurtz]

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

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

That is what you need to prove.

 

[frenchkiki]

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:
$\forall$ ε>0, ∃ N₁>0, ∀x, x<N₁→ L-ε<g(x)<L+ε

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

Let N=max(N₁,N₂)

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

therefore lim g(x)=L as x→-∞

 

[LCKurtz]

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:
$\forall$ ε>0, ∃ N₁>0, ∀x, x<N₁→ L-ε<g(x)<L+ε

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

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.

Let N=max(N₁,N₂)

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

therefore lim g(x)=L as x→-∞

Fix that with a couple of minor changes and you will be good to go.

 

[frenchkiki]

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→-∞

 

[LCKurtz]

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→-∞

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.

 

[frenchkiki]

OK. Thanks a lot for your help LCKurtz.