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3.6: Limit Superior and Limit Inferior of Functions

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    49111
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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

    Add proof here and it will automatically be hidden

    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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    This page titled 3.6: Limit Superior and Limit Inferior of Functions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Lafferriere, Lafferriere, and Nguyen (PDXOpen: Open Educational Resources) .

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