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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