# 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.

## Definition $$\PageIndex{1}$$

Let $$f: E \rightarrow \mathbb{R}$$ and let $$\bar{x}$$ be a limit point of $$D$$. Recall that

$B_{0}(\bar{x} ; \delta)=B_{-}(\bar{x} ; \delta) \cup B_{+}(\bar{x} ; \delta)=(\bar{x}-\delta, \bar{x}) \cup(\bar{x}, \bar{x}+\delta) .$

The limit superior of the function $$f$$ at $$\bar{x}$$ is defnied by

$\limsup _{x \rightarrow \bar{x}} f(x)=\inf _{\delta>0} \sup _{x \in B_{0}(\bar{x} ; \delta) \cap D} f(x) .$

Similarly, the limit inferior of the function $$f$$ at $$\bar{x}$$ is defineid by

$\liminf _{x \rightarrow \bar{x}} f(x)=\sup _{\delta>0} \inf _{x \in B_{0}(\bar{x} ; \delta) \cap D} f(x) .$

Consider the extended real-valued function $$g:(0, \infty) \rightarrow(-\infty, \infty]$$ defined by

$g(\delta)=\sup _{x \in B_{0}(\vec{x} ; \delta) \cap D} f(x)$

It is clear that $$g$$ is increasing and

$\limsup _{x \rightarrow \bar{x}} f(x)=\inf _{\delta>0} g(\delta) .$

We say that the function $$f$$ is locally bounded above around $$\bar{x}$$ if there exists $$\delta > 0$$ and $$M > 0$$ such that

$f(x) \leq M \text { for all } x \in B(\bar{x} ; \delta) \cap D .$

Clearly, if $$f$$ is locally bounded above around $$\bar{x}$$, then $$\limsup _{x \rightarrow \bar{x}} f(x)$$ is a real number, while $$\limsup _{x \rightarrow z} f(x)=\infty$$ in the other case. Similar discussion applies for the limit inferior.

## Theorem $$\PageIndex{1}$$

Let $$f: D \rightarrow \mathbb{R}$$ and let $$\bar{x}$$ be a limit point of $$D$$. Then $$\ell=\limsup _{x \rightarrow \bar{x}} f(x)$$ if and only if the following two conditions hold:

1. For every $$\varepsilon > 0$$, there exists $$\delta > 0$$ such that

$f(x)<\ell+\varepsilon \text { for all } x \in B_{0}(\bar{x} ; \delta) \cap D ;$

1. For every $$\varepsilon > 0$$ and for every $$\delta > 0$$, there exists $$x_{\delta} \in B_{0}(\bar{x} ; \delta) \cap D$$ such that

$\ell-\varepsilon<f\left(x_{\delta}\right)$

Proof

Suppose $$\ell=\limsup _{x \rightarrow \bar{x}} f(x)$$. Then

$\ell=\inf _{\delta>0} g(\delta) ,$

where $$g$$ is defined in (3.10). For any $$\varepsilon > 0$$, there exists $$\delta > 0$$ such that

$\ell \leq g(\delta)=\sup _{x \in B_{0}(\bar{x} ; \delta) \cap D} f(x)<\ell+\varepsilon .$

Thus,

$f(x)<\ell+\varepsilon \text { for all } x \in B_{0}(\bar{x} ; \delta) \cap D ,$

which proves conditions (1). Next note that for any $$\varepsilon > 0$$ and $$\delta > 0$$, we have

$\ell-\varepsilon<\ell \leq g(\delta)=\sup _{x \in B_{0}(\bar{x} ; \delta) \cap D} f(x) .$

Thus, there exists $$x_{\delta} \in B_{0}(\bar{x} ; \delta) \cap D$$ with

$\ell-\varepsilon<f\left(x_{\delta}\right) .$

This proves (2).

Let us now prove the converse. Suppose (1) and (2) are satisfied. Fix any $$\varepsilon > 0$$ and let $$\delta > 0$$ satisfy (1). Then

$g(\delta)=\sup _{x \in B_{0}(\bar{x} ; \delta) \cap D} f(x) \leq \ell+\varepsilon .$

This implies

$\limsup _{x \rightarrow \bar{x}} f(x)=\inf _{\delta>0} g(\delta) \leq \ell+\varepsilon .$

Since $$\varepsilon$$ is arbitrary, we get

$\limsup _{x \rightarrow \bar{x}} f(x) \leq \ell .$

Again, let $$\varepsilon > 0$$. Given $$\delta > 0$$, let $$x_{\delta}$$ be as in (2). Therefore,

$\ell-\varepsilon<f\left(x_{\delta}\right) \leq \sup _{x \in B_{0}(\bar{x} ; \delta) \cap D} f(x)=g(\delta) .$

This implies

$\ell-\varepsilon \leq \inf _{\delta>0} g(\delta)=\limsup _{x \rightarrow \bar{x}} f(x) .$

It follows that $$\ell \leq \limsup _{x \rightarrow \bar{x}} f(x)$$. Therefore, $$\ell=\limsup _{x \rightarrow \bar{x}} f(x)$$. $$\square$$

## Corollary $$\PageIndex{2}$$

Suppose $$\ell=\limsup _{x \rightarrow \bar{x}} f(x)$$. Then there exists a sequence $$\left\{x_{k}\right\}$$ in $$D$$ such that $$\left\{x_{k}\right\}$$ converges to $$\bar{x}$$, $$x_{k} \neq \bar{x}$$ for every $$k$$, and

$\lim _{k \rightarrow \infty} f\left(x_{k}\right)=\ell .$

Moreover, if $$\left\{y_{k}\right\}$$ is a sequence in $$D$$ that converges to $$\bar{x}$$, $$y_{k} \neq \bar{x}$$ for every $$k$$, and $$\lim _{k \rightarrow \infty} f\left(y_{k}\right)=\ell^{\prime}$$, then $$\ell^{\prime} \leq \ell$$.

Proof

Let $$\delta_{k}^{\prime}=\min \left\{\delta_{k}, \frac{1}{k}\right\}$$. Then $$\delta_{k}^{\prime} \leq \delta_{k}$$ and $$\lim _{k \rightarrow \infty} \delta_{k}^{\prime}=0$$. From (2) of Theorem 3.6.1, there exists $$x_{k} \in B_{0}\left(\bar{x} ; \delta_{k}^{\prime}\right) \cap D$$ such that

$\ell-\varepsilon_{k}<f\left(x_{k}\right) .$

Moreover, $$f\left(x_{k}\right)<\ell+\varepsilon_{k}$$ by (3.11). Therefore, $$\left\{x_{k}\right\}$$ is a sequence that satisfies the conclusion of the corollary.

Now let $$\left\{y_{k}\right\}$$ be a sequence in $$D$$ that converges to $$\bar{x}$$, $$y_{k} \neq \bar{x}$$ for every $$k$$, and $$\lim _{k \rightarrow \infty} f\left(y_{k}\right)=\ell^{\prime}$$. For any $$\varepsilon > 0$$, let $$\delta > 0$$ be as in (1) of Theorem 3.6.1. Since $$y_{k} \in B_{0}(\bar{x} ; \delta) \cap D$$ when $$k$$ is sufficiently large, we have

$f\left(y_{k}\right)<\ell+\varepsilon$

for such $$k$$. This implies $$\ell^{\prime} \leq \ell+\varepsilon$$. It follows that $$\ell^{\prime} \leq \ell$$. $$\square$$

## Remark 3.6.3

Let $$f: D \rightarrow \mathbb{R}$$ and let $$\bar{x}$$ be a limit point of $$D$$. Suppose $$\limsup _{x \rightarrow \bar{x}} f(x)$$ is a real number. Define

$A=\left\{\ell \in \mathbb{R}: \exists\left\{x_{k}\right\} \subset D, x_{k} \neq \bar{x} \text { for every } k, x_{k} \rightarrow \bar{x}, f\left(x_{k}\right) \rightarrow \ell\right\} .$

Then the previous corollary shows that $$A \neq \emptyset$$ and $$\limsup _{x \rightarrow \bar{x}} f(x)=\max A$$.

## Theorem $$\PageIndex{4}$$

Let $$f: D \rightarrow \mathbb{R}$$ and let $$\bar{x}$$ be a limit point of $$D$$. Then

$\limsup _{x \rightarrow \bar{x}} f(x)=\infty$

if and only if there exists a sequence $$\left\{x_{k}\right\}$$ in $$D$$ such that $$\left\{x_{k}\right\}$$ converges to $$\bar{x}$$, $$x_{k} \neq \bar{x}$$ for every $$k$$, and $$\lim _{k \rightarrow \infty} f\left(x_{k}\right)=\infty$$.

Proof

Suppose $$\limsup _{x \rightarrow \bar{x}} f(x)=\infty$$. Then

$\inf _{\delta>0} g(\delta)=\infty ,$

wehre $$g$$ is the extended real-valued function defined in (3.10). Thus, $$g(\boldsymbol{\delta})=\infty$$ for every $$\delta > 0$$. Given $$k \in \mathbb{N}$$, for $$\delta_{k}=\frac{1}{k}$$, since

$g\left(\delta_{k}\right)=\sup _{x \in B_{0}\left(\bar{x} ; \delta_{k}\right) \cap D} f(x)=\infty ,$

there exists $$x_{k} \in B_{0}\left(\bar{x} ; \delta_{k}\right) \cap D$$ such that $$f\left(x_{k}\right)>k$$. Therefore, $$\lim _{k \rightarrow \infty} f\left(x_{k}\right)=\infty$$.

Let us prove the converse. Since $$\lim _{k \rightarrow \infty} f\left(x_{k}\right)=\infty$$, for every $$M \in \mathbb{R}$$, there exists $$K \in \mathbb{N}$$ such that

$f\left(x_{k}\right) \geq M \text { for every } k \geq K .$

For any $$\delta > 0$$, we have

$x_{k} \in B_{0}(\bar{x} ; \delta) \cap D .$

This implies $$g(\delta)=\infty$$, and hence $$\limsup _{x \rightarrow \bar{x}} f(x)=\infty$$. $$\square$$

## Theorem $$\PageIndex{5}$$

Let $$f: D \rightarrow \mathbb{R}$$ and let $$\bar{x}$$ be a limit point of $$D$$. Then

$\limsup _{x \rightarrow \bar{x}} f(x)=-\infty$

if and only if for any sequence $$\left\{x_{k}\right\}$$ in $$D$$ such that $$\left\{x_{k}\right\}$$ converges to $$\bar{x}$$, $$x_{k} \neq \bar{x}$$ for every $$k$$, it follows that $$\lim _{k \rightarrow \infty} f\left(x_{k}\right)=-\infty$$. The latter is equivalent to $$\lim _{x \rightarrow \bar{x}} f(x)=-\infty$$.

Following the same arguments, we can prove similar results for inferior limits of functions.

Proof

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## Theorem $$\PageIndex{6}$$

Let $$f: D \rightarrow \mathbb{R}$$ and let $$\bar{x}$$ be a limit point of $$D$$. Then $$\ell=\liminf _{x \rightarrow \bar{x}} f(x)$$ if and only if the following two conditions hold:

1. For every $$\varepsilon > 0$$, there exists $$\delta > 0$$ such that

$\ell-\varepsilon<f(x) \text { for all } x \in B_{0}(\bar{x} ; \delta) \cap D ;$

1. For every $$\varepsilon > 0$$ and for every $$\delta > 0$$, there exists $$x \in B_{0}(\bar{x} ; \delta) \cap D$$ such that

$f(x)<\ell+\varepsilon .$

Proof

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## Theorem $$\PageIndex{7}$$

Suppose $$\ell=\liminf _{x \rightarrow \bar{x}} f(x)$$. Then there exists a sequence $$\left\{x_{k}\right\}$$ in $$D$$ such that $$x_{k}$$ converges to $$\bar{x}$$, $$x_{k} \neq \bar{x}$$ for every $$k$$, and

$\lim _{k \rightarrow \infty} f\left(x_{k}\right)=\ell .$

Moreover, if $$\left\{y_{k}\right\}$$ is a sequence in $$D$$ that converges to $$\bar{x}$$, $$y_{k} \neq \bar{x}$$ for every $$k$$, and $$\lim _{k \rightarrow \infty} f\left(y_{k}\right)=\ell^{\prime}$$, then $$\ell^{\prime} \geq \ell$$.

Proof

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## Remark $$\PageIndex{8}$$

Let $$f: D \rightarrow \mathbb{R}$$ and let $$\bar{x}$$ be a limit point of $$D$$. Suppose $$\liminf _{x \rightarrow \bar{x}} f(x)$$ is a real number. Define

$B=\left\{\ell \in \mathbb{R}: \exists\left\{x_{k}\right\} \subset D, x_{k} \neq \bar{x} \text { for every } \mathrm{k}, x_{k} \rightarrow \bar{x}, f\left(x_{k}\right) \rightarrow \ell\right\} .$

Then $$B \neq \emptyset$$ and $$\liminf _{x \rightarrow \bar{x}} f(x)=\min B$$.

## Theorem $$\PageIndex{9}$$

Let $$f: D \rightarrow \mathbb{R}$$ and let $$\bar{x}$$ be a limit point of $$D$$. Then

$\liminf _{x \rightarrow \bar{x}} f(x)=-\infty$

if and only if there exists a sequence $$\left\{x_{k}\right\}$$ in $$D$$ such that $$\left\{x_{k}\right\}$$ converges to $$\bar{x}$$, $$x_{k} \neq \bar{x}$$ for every $$k$$, and $$\lim _{k \rightarrow \infty} f\left(x_{k}\right)=-\infty$$.

Proof

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## Theorem $$\PageIndex{10}$$

Let $$f: D \rightarrow \mathbb{R}$$ and let $$\bar{x}$$ be a limit point of $$D$$. Then

$\liminf _{x \rightarrow \bar{x}} f(x)=-\infty$

if and only if there exists a sequence $$\left\{x_{k}\right\}$$ in $$D$$ such that $$\left\{x_{k}\right\}$$ converges to $$\bar{x}$$, $$x_{k} \neq \bar{x}$$ for every $$k$$, it follows $$\lim _{k \rightarrow \infty} f\left(x_{k}\right)=\infty$$. The latter is equivalent to $$\lim _{x \rightarrow \bar{x}} f(x)=\infty$$.

Proof

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## Theorem $$\PageIndex{11}$$

Let $$f: D \rightarrow \mathbb{R}$$ and let $$\bar{x}$$ be a limit point of $$D$$. Then

$\lim _{x \rightarrow \bar{x}} f(x)=\ell .$

if and only if

$\limsup _{x \rightarrow \bar{x}} f(x)=\liminf _{x \rightarrow \bar{x}} f(x)=\ell .$

Proof

Suppose

$\lim _{x \rightarrow \bar{x}} f(x)=\ell .$

Then for every $$\varepsilon > 0$$, there exists $$\delta > 0$$ such that

$\ell-\varepsilon<f(x)<\ell+\varepsilon \text { for all } x \in B_{0}(\bar{x} ; \delta) \cap D .$

Since this also holds for every $$0<\delta^{\prime}<\delta$$, we get

$\ell-\varepsilon<g\left(\delta^{\prime}\right) \leq \ell+\varepsilon .$

It follows that

$\ell-\varepsilon \leq \inf _{\delta^{\prime}>0} g\left(\delta^{\prime}\right) \leq \ell+\varepsilon .$

Therefore, $$\limsup _{x \rightarrow \bar{x}} f(x)=\ell$$ since $$\varepsilon$$ 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. $$\square$$

Exercise $$\PageIndex{1}$$

Let $$D \subset \mathbb{R}$$, $$f: D \rightarrow \mathbb{R}$$, and $$\bar{x}$$ be a limit point of $$D$$. Prove that $$\liminf _{x \rightarrow \bar{x}} f(x) \leq \limsup _{x \rightarrow \bar{x}} f(x)$$.

Add texts here. Do not delete this text first.

Exercise $$\PageIndex{2}$$

Find each of the following limits:

1. $$\limsup _{x \rightarrow 0} \sin \left(\frac{1}{x}\right)$$.
2. $$\liminf _{x \rightarrow 0} \sin \left(\frac{1}{x}\right)$$.
3. $$\limsup _{x \rightarrow 0} \frac{\cos x}{x}$$.
4. $$\liminf _{x \rightarrow 0} \frac{\cos x}{x}$$.