3.6: Limit Superior and Limit Inferior of Functions
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We extend to functions and concepts of limit superior and limit inferior.
Let f:E→R and let ˉx be a limit point of D. Recall that
B0(ˉx;δ)=B−(ˉx;δ)∪B+(ˉx;δ)=(ˉx−δ,ˉx)∪(ˉx,ˉx+δ).
The limit superior of the function f at ˉx is defnied by
lim supx→ˉxf(x)=infδ>0supx∈B0(ˉx;δ)∩Df(x).
Similarly, the limit inferior of the function f at ˉx is defineid by
lim infx→ˉxf(x)=supδ>0infx∈B0(ˉx;δ)∩Df(x).
Consider the extended real-valued function g:(0,∞)→(−∞,∞] defined by
g(δ)=supx∈B0(→x;δ)∩Df(x)
It is clear that g is increasing and
lim supx→ˉxf(x)=infδ>0g(δ).
We say that the function f is locally bounded above around ˉx if there exists δ>0 and M>0 such that
f(x)≤M for all x∈B(ˉx;δ)∩D.
Clearly, if f is locally bounded above around ˉx, then lim supx→ˉxf(x) is a real number, while lim supx→zf(x)=∞ in the other case. Similar discussion applies for the limit inferior.
Let f:D→R and let ˉx be a limit point of D. Then ℓ=lim supx→ˉxf(x) if and only if the following two conditions hold:
- For every ε>0, there exists δ>0 such that
f(x)<ℓ+ε for all x∈B0(ˉx;δ)∩D;
- For every ε>0 and for every δ>0, there exists xδ∈B0(ˉx;δ)∩D such that
ℓ−ε<f(xδ)
- Proof
-
Suppose ℓ=lim supx→ˉxf(x). Then
ℓ=infδ>0g(δ),
where g is defined in (3.10). For any ε>0, there exists δ>0 such that
ℓ≤g(δ)=supx∈B0(ˉx;δ)∩Df(x)<ℓ+ε.
Thus,
f(x)<ℓ+ε for all x∈B0(ˉx;δ)∩D,
which proves conditions (1). Next note that for any ε>0 and δ>0, we have
ℓ−ε<ℓ≤g(δ)=supx∈B0(ˉx;δ)∩Df(x).
Thus, there exists xδ∈B0(ˉx;δ)∩D with
ℓ−ε<f(xδ).
This proves (2).
Let us now prove the converse. Suppose (1) and (2) are satisfied. Fix any ε>0 and let δ>0 satisfy (1). Then
g(δ)=supx∈B0(ˉx;δ)∩Df(x)≤ℓ+ε.
This implies
lim supx→ˉxf(x)=infδ>0g(δ)≤ℓ+ε.
Since ε is arbitrary, we get
lim supx→ˉxf(x)≤ℓ.
Again, let ε>0. Given δ>0, let xδ be as in (2). Therefore,
ℓ−ε<f(xδ)≤supx∈B0(ˉx;δ)∩Df(x)=g(δ).
This implies
ℓ−ε≤infδ>0g(δ)=lim supx→ˉxf(x).
It follows that ℓ≤lim supx→ˉxf(x). Therefore, ℓ=lim supx→ˉxf(x). ◻
Suppose ℓ=lim supx→ˉxf(x). Then there exists a sequence {xk} in D such that {xk} converges to ˉx, xk≠ˉx for every k, and
limk→∞f(xk)=ℓ.
Moreover, if {yk} is a sequence in D that converges to ˉx, yk≠ˉx for every k, and limk→∞f(yk)=ℓ′, then ℓ′≤ℓ.
- Proof
-
Let δ′k=min{δk,1k}. Then δ′k≤δk and limk→∞δ′k=0. From (2) of Theorem 3.6.1, there exists xk∈B0(ˉx;δ′k)∩D such that
ℓ−εk<f(xk).
Moreover, f(xk)<ℓ+εk by (3.11). Therefore, {xk} is a sequence that satisfies the conclusion of the corollary.
Now let {yk} be a sequence in D that converges to ˉx, yk≠ˉx for every k, and limk→∞f(yk)=ℓ′. For any ε>0, let δ>0 be as in (1) of Theorem 3.6.1. Since yk∈B0(ˉx;δ)∩D when k is sufficiently large, we have
f(yk)<ℓ+ε
for such k. This implies ℓ′≤ℓ+ε. It follows that ℓ′≤ℓ. ◻
Let f:D→R and let ˉx be a limit point of D. Suppose lim supx→ˉxf(x) is a real number. Define
A={ℓ∈R:∃{xk}⊂D,xk≠ˉx for every k,xk→ˉx,f(xk)→ℓ}.
Then the previous corollary shows that A≠∅ and lim supx→ˉxf(x)=maxA.
Let f:D→R and let ˉx be a limit point of D. Then
lim supx→ˉxf(x)=∞
if and only if there exists a sequence {xk} in D such that {xk} converges to ˉx, xk≠ˉx for every k, and limk→∞f(xk)=∞.
- Proof
-
Suppose lim supx→ˉxf(x)=∞. Then
infδ>0g(δ)=∞,
wehre g is the extended real-valued function defined in (3.10). Thus, g(δ)=∞ for every δ>0. Given k∈N, for δk=1k, since
g(δk)=supx∈B0(ˉx;δk)∩Df(x)=∞,
there exists xk∈B0(ˉx;δk)∩D such that f(xk)>k. Therefore, limk→∞f(xk)=∞.
Let us prove the converse. Since limk→∞f(xk)=∞, for every M∈R, there exists K∈N such that
f(xk)≥M for every k≥K.
For any δ>0, we have
xk∈B0(ˉx;δ)∩D.
This implies g(δ)=∞, and hence lim supx→ˉxf(x)=∞. ◻
Let f:D→R and let ˉx be a limit point of D. Then
lim supx→ˉxf(x)=−∞
if and only if for any sequence {xk} in D such that {xk} converges to ˉx, xk≠ˉx for every k, it follows that limk→∞f(xk)=−∞. The latter is equivalent to limx→ˉxf(x)=−∞.
Following the same arguments, we can prove similar results for inferior limits of functions.
- Proof
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Let f:D→R and let ˉx be a limit point of D. Then ℓ=lim infx→ˉxf(x) if and only if the following two conditions hold:
- For every ε>0, there exists δ>0 such that
ℓ−ε<f(x) for all x∈B0(ˉx;δ)∩D;
- For every ε>0 and for every δ>0, there exists x∈B0(ˉx;δ)∩D such that
f(x)<ℓ+ε.
- Proof
-
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Suppose ℓ=lim infx→ˉxf(x). Then there exists a sequence {xk} in D such that xk converges to ˉx, xk≠ˉx for every k, and
limk→∞f(xk)=ℓ.
Moreover, if {yk} is a sequence in D that converges to ˉx, yk≠ˉx for every k, and limk→∞f(yk)=ℓ′, then ℓ′≥ℓ.
- Proof
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Add proof here and it will automatically be hidden
Let f:D→R and let ˉx be a limit point of D. Suppose lim infx→ˉxf(x) is a real number. Define
B={ℓ∈R:∃{xk}⊂D,xk≠ˉx for every k,xk→ˉx,f(xk)→ℓ}.
Then B≠∅ and lim infx→ˉxf(x)=minB.
Let f:D→R and let ˉx be a limit point of D. Then
lim infx→ˉxf(x)=−∞
if and only if there exists a sequence {xk} in D such that {xk} converges to ˉx, xk≠ˉx for every k, and limk→∞f(xk)=−∞.
- Proof
-
Add proof here and it will automatically be hidden
Let f:D→R and let ˉx be a limit point of D. Then
lim infx→ˉxf(x)=−∞
if and only if there exists a sequence {xk} in D such that {xk} converges to ˉx, xk≠ˉx for every k, it follows limk→∞f(xk)=∞. The latter is equivalent to limx→ˉxf(x)=∞.
- Proof
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Add proof here and it will automatically be hidden
Let f:D→R and let ˉx be a limit point of D. Then
limx→ˉxf(x)=ℓ.
if and only if
lim supx→ˉxf(x)=lim infx→ˉxf(x)=ℓ.
- Proof
-
Suppose
limx→ˉxf(x)=ℓ.
Then for every ε>0, there exists δ>0 such that
ℓ−ε<f(x)<ℓ+ε for all x∈B0(ˉx;δ)∩D.
Since this also holds for every 0<δ′<δ, we get
ℓ−ε<g(δ′)≤ℓ+ε.
It follows that
ℓ−ε≤infδ′>0g(δ′)≤ℓ+ε.
Therefore, lim supx→ˉxf(x)=ℓ since ε is arbitrary. The proof for the limit inferior is similar. The converse follows directly from (1) of Theorem 3.6.1 and Theorem 3.6.6. ◻
Exercise 3.6.1
Let D⊂R, f:D→R, and ˉx be a limit point of D. Prove that lim infx→ˉxf(x)≤lim supx→ˉxf(x).
- Answer
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Exercise 3.6.2
Find each of the following limits:
- lim supx→0sin(1x).
- lim infx→0sin(1x).
- lim supx→0cosxx.
- lim infx→0cosxx.
- Answer
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