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Section 4.2 Limits and Continuity

Subsection 4.2.1 Limits at Cluster Points

Let \(S \subseteq \R\text{,}\) let \(f \colon S \to \R\) be a function, and let \(c\) be a cluster point of \(S\text{.}\) We say that \(f\) has limit\(L \in \R\) at \(c\) if for every \(\varepsilon \gt 0\) there exists \(\delta \gt 0\) such that whenever \(x \in S\) and \(0 \lt |x-c| \lt \delta\text{,}\) we have
\begin{equation*} |f(x)-L| \lt \varepsilon. \end{equation*}
In that case we write
\begin{equation*} \lim_{x\to c} f(x)=L \end{equation*}
or, equivalently,
\begin{equation*} f(x)\to L \quad \text{as } x\to c. \end{equation*}
These two notations mean exactly the same thing. The second is read as “\(f(x)\) tends to \(L\) as \(x\) tends to \(c\text{.}\)”
The condition \(0 \lt |x-c|\) is essential: it says that we are only looking at points \(x\) near \(c\) but different from \(c\text{.}\) Thus the existence of \(\lim_{x\to c} f(x)\) depends only on the values of \(f\) arbitrarily close to \(c\text{,}\) not on the value \(f(c)\text{.}\) In particular, \(c\) need not even belong to the domain of \(f\text{.}\)

Example 4.2.1.

Let \(f(x)=x+1\) on \(\R \setminus \{1\}\text{.}\) Then \(\lim_{x\to 1} f(x)=2\text{.}\) Indeed, if \(0 \lt |x-1| \lt \varepsilon\text{,}\) then
\begin{equation*} |f(x)-2|=|(x+1)-2|=|x-1| \lt \varepsilon. \end{equation*}
This example shows that a limit may exist even when \(f\) is not defined at the limiting point.

Proof.
Suppose \(L_1\) and \(L_2\) are limits of \(f\) at \(c\text{.}\) We must show that \(L_1=L_2\text{.}\) Let \(\varepsilon \gt 0\text{.}\) By the definition of limit, there exist \(\delta_i \gt 0\) such that \(|f(x)-L_i| \lt \varepsilon\) whenever \(x \in S\) and \(0 \lt |x-c| \lt \delta_i\text{,}\) for \(i=1,2\text{.}\) Thus,
\begin{equation*} |L_1-L_2| \le |f(x)-L_1| + |f(x)-L_2| \lt 2\varepsilon \end{equation*}
for any \(x \in S\) with \(0 \lt |x-c| \lt \min\{\delta_1,\delta_2\}\text{.}\) By the definition of cluster point, such an \(x\) exists. Since \(\varepsilon \gt 0\) was arbitrary, it follows that \(|L_1-L_2|=0\text{.}\) Therefore \(L_1=L_2\text{.}\)
This proposition also explains why we only define \(\lim_{x\to c} f(x)\) when \(c\) is a cluster point of the domain. If \(c\) were an isolated point of \(S\text{,}\) then for sufficiently small \(\delta\) there would be no point \(x \in S\) with \(0 \lt |x-c| \lt \delta\text{.}\) The implication in the definition would then be automatically true for every real number \(L\text{,}\) so the “limit” would not be unique.

Example 4.2.3.

Let \(S=\{0\}\) and let \(f(0)=7\text{.}\) If we tried to define a limit of \(f\) at \(0\) using the usual punctured-neighborhood condition, then every \(L \in \R\) would satisfy it, because there are no points \(x \in S\) with \(0 \lt |x| \lt \delta\text{.}\) So there is no meaningful notion of limit at that isolated point.

Subsection 4.2.2 Left and Right Limits

Suppose \(c\) is a cluster point of \(S \cap (-\infty,c)\text{.}\) We say that \(f\) has left limit\(L\) at \(c\) if for every \(\varepsilon \gt 0\) there exists \(\delta \gt 0\) such that whenever \(x \in S\) and \(c-\delta \lt x \lt c\text{,}\) we have
\begin{equation*} |f(x)-L| \lt \varepsilon. \end{equation*}
We write \(\lim_{x\to c^-} f(x)=L\) or \(f(c^-) = L\text{.}\)
Similarly, if \(c\) is a cluster point of \(S \cap (c,\infty)\text{,}\) we say that \(f\) has right limit\(L\) at \(c\) if for every \(\varepsilon \gt 0\) there exists \(\delta \gt 0\) such that whenever \(x \in S\) and \(c \lt x \lt c+\delta\text{,}\) we have
\begin{equation*} |f(x)-L| \lt \varepsilon. \end{equation*}
We write \(\lim_{x\to c^+} f(x)=L\) or \(f(c^+) = L\text{.}\)

Proof.
Suppose \(\lim_{x\to c} f(x)=L\text{.}\) Given \(\varepsilon \gt 0\text{,}\) choose \(\delta \gt 0\) so that \(|f(x)-L| \lt \varepsilon\) whenever \(x \in S\) and \(0 \lt |x-c| \lt \delta\text{.}\) If \(c-\delta \lt x \lt c\text{,}\) then certainly \(0 \lt |x-c| \lt \delta\text{,}\) so \(|f(x)-L| \lt \varepsilon\text{.}\) This proves \(\lim_{x\to c^-} f(x)=L\text{.}\) The proof for the right limit is the same.
Conversely, suppose both one-sided limits exist and are equal to \(L\text{.}\) Given \(\varepsilon \gt 0\text{,}\) choose \(\delta_1,\delta_2 \gt 0\) so that \(|f(x)-L| \lt \varepsilon\) whenever \(x \in S\) and \(c-\delta_1 \lt x \lt c\text{,}\) and also whenever \(x \in S\) and \(c \lt x \lt c+\delta_2\text{.}\) Let \(\delta=\min\{\delta_1,\delta_2\}\text{.}\) If \(x \in S\) and \(0 \lt |x-c| \lt \delta\text{,}\) then either \(x \lt c\) or \(x \gt c\text{.}\) In the first case \(c-\delta_1 \lt x \lt c\text{,}\) and in the second case \(c \lt x \lt c+\delta_2\text{.}\) Hence in either case \(|f(x)-L| \lt \varepsilon\text{.}\) Therefore \(\lim_{x\to c} f(x)=L\text{.}\)

Example 4.2.5.

Define
\begin{equation*} \operatorname{sgn} x=\begin{cases} -1 & \text{if } x \lt 0,\\ 1 & \text{if } x \ge 0. \end{cases} \end{equation*}
Then \(\lim_{x\to 0^-} \operatorname{sgn} x=-1\) and \(\lim_{x\to 0^+} \operatorname{sgn} x=1\text{.}\) Since these one-sided limits are different, \(\lim_{x\to 0} \operatorname{sgn} x\) does not exist.

Subsection 4.2.3 Relation Between Limits and Continuity

The definition of limit ignores the value at the point \(c\text{,}\) whereas continuity compares nearby values with \(f(c)\text{.}\) Because of this, limits and continuity are closely related but not identical.

Proof.
Suppose \(f\) is continuous at \(c\text{.}\) Let \(\varepsilon \gt 0\) be given. By continuity, there exists \(\delta \gt 0\) such that \(|f(x)-f(c)| \lt \varepsilon\) whenever \(x \in S\) and \(|x-c| \lt \delta\text{.}\) In particular, this holds whenever \(x \in S\) and \(0 \lt |x-c| \lt \delta\text{.}\) Hence \(\lim_{x\to c} f(x)=f(c)\text{.}\)
Conversely, suppose \(\lim_{x\to c} f(x)=f(c)\text{.}\) Let \(\varepsilon \gt 0\) be given. Then there exists \(\delta \gt 0\) such that \(|f(x)-f(c)| \lt \varepsilon\) whenever \(x \in S\) and \(0 \lt |x-c| \lt \delta\text{.}\) Now if \(x \in S\) and \(|x-c| \lt \delta\text{,}\) then either \(x=c\text{,}\) in which case \(|f(x)-f(c)|=0\text{,}\) or \(0 \lt |x-c| \lt \delta\text{,}\) in which case \(|f(x)-f(c)| \lt \varepsilon\text{.}\) Therefore \(f\) is continuous at \(c\text{.}\)

Example 4.2.7.

Define
\begin{equation*} f(x)=\begin{cases} \dfrac{x^2-1}{x-1} & \text{if } x \neq 1,\\ 0 & \text{if } x=1. \end{cases} \end{equation*}
For \(x \neq 1\text{,}\) we have \(f(x)=x+1\text{,}\) so \(\lim_{x\to 1} f(x)=2\text{.}\) However, \(f(1)=0\text{,}\) so \(f\) is not continuous at \(1\text{.}\) If we redefine the value at \(1\) to be \(2\text{,}\) then the resulting function becomes continuous at \(1\text{.}\) This is the basic model of a removable discontinuity.

Subsection 4.2.4 Limits Involving \(\pm\infty\)

We now extend the definition of limit to cases where either the point approached or the limiting value is allowed to be \(+\infty\) or \(-\infty\text{.}\) The set \(\R \cup \{-\infty,+\infty\}\) is called the extended real line.
First suppose \(c \in \R\) is a cluster point of \(S\text{.}\) We write
\begin{equation*} \lim_{x\to c} f(x)=+\infty \end{equation*}
if for every \(M \in \R\) there exists \(\delta \gt 0\) such that whenever \(x \in S\) and \(0 \lt |x-c| \lt \delta\text{,}\) we have \(f(x) \gt M\text{.}\) Similarly, we write
\begin{equation*} \lim_{x\to c} f(x)=-\infty \end{equation*}
if for every \(M \in \R\) there exists \(\delta \gt 0\) such that whenever \(x \in S\) and \(0 \lt |x-c| \lt \delta\text{,}\) we have \(f(x) \lt M\text{.}\)
The one-sided definitions extend in exactly the same way. For example, \(\lim_{x\to c^+} f(x)=+\infty\) means that for every \(M \in \R\) there exists \(\delta \gt 0\) such that whenever \(x \in S\) and \(c \lt x \lt c+\delta\text{,}\) we have \(f(x) \gt M\text{.}\) The other one-sided cases are analogous.
Next suppose that \(S\) is unbounded above. We write
\begin{equation*} \lim_{x\to +\infty} f(x)=L \end{equation*}
for \(L \in \R\) if for every \(\varepsilon \gt 0\) there exists \(A \in \R\) such that whenever \(x \in S\) and \(x \gt A\text{,}\) we have \(|f(x)-L| \lt \varepsilon\text{.}\) We write
\begin{equation*} \lim_{x\to +\infty} f(x)=+\infty \end{equation*}
if for every \(M \in \R\) there exists \(A \in \R\) such that whenever \(x \in S\) and \(x \gt A\text{,}\) we have \(f(x) \gt M\text{.}\) The definition of \(\lim_{x\to +\infty} f(x)=-\infty\) is analogous.
Likewise, if \(S\) is unbounded below, we write
\begin{equation*} \lim_{x\to -\infty} f(x)=L \end{equation*}
for \(L \in \R\) if for every \(\varepsilon \gt 0\) there exists \(A \in \R\) such that whenever \(x \in S\) and \(x \lt A\text{,}\) we have \(|f(x)-L| \lt \varepsilon\text{.}\) We define \(\lim_{x\to -\infty} f(x)=+\infty\) and \(\lim_{x\to -\infty} f(x)=-\infty\) similarly, by requiring \(f(x)\) to be eventually larger than any prescribed real number, or eventually smaller than any prescribed real number.
In each of these cases we also use the corresponding arrow notation: \(f(x)\to L\) as \(x\to c\text{,}\) as \(x\to +\infty\text{,}\) or as \(x\to -\infty\text{.}\) The same uniqueness principle remains valid for these extended-real limits as well.

Example 4.2.8.

  • On \(\R \setminus \{0\}\text{,}\) we have \(\lim_{x\to 0} \frac{1}{x^2}=+\infty\text{.}\)
  • On \(\R \setminus \{0\}\text{,}\) we have \(\lim_{x\to 0^+} \frac{1}{x}=+\infty\) and \(\lim_{x\to 0^-} \frac{1}{x}=-\infty\text{.}\)
  • On \((0,\infty)\text{,}\) we have \(\lim_{x\to +\infty} \frac{1}{x}=0\text{.}\)
  • On \(\R\text{,}\) we have \(\lim_{x\to -\infty} x^2=+\infty\text{.}\)
It is often helpful to think of \(+\infty\) and \(-\infty\) as extra endpoints added to the real line. Then the definitions above are the natural analogues of ordinary limits at finite cluster points.
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