3.6: Limit Superior and Limit Inferior of Functions

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

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

2. 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

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

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

Answer

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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. $\liminf _{x \rightarrow 0} \frac{\cos x}{x}$.

Answer

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