2.1 Limits of functions
Definition 2.1.1. Let \(X\) and \(Y\) be subsets of \(\mathbb {R}\). Suppose \(E\subseteq X\), \(f\) maps \(E\) into \(Y\), and \(p\) is a limit point of \(E\). We say that \(f(x)\) tends to \(q\) as \(x\) approaches \(p\) in \(E\), and write \[\lim _{x\rightarrow p} f(x) = q,\] if for every \(\varepsilon >0\) there exists \(\delta > 0\) such that \(|f(x) - q| < \varepsilon \) for every \(x \in E\) with \(0 < |x - p| < \delta \).
This definition can be recast in terms of limits of sequences as follows.
Theorem 2.1.2. Let \(X\) and \(Y\) be subsets of \(\mathbb {R}\). Suppose \(E\subseteq X\), \(f\) maps \(E\) into \(Y\), and \(p\) is a limit point of \(E\). Then \begin {equation} \label {eq:limfx} \lim _{x\rightarrow p} f(x) = q \end {equation} if and only if \begin {equation} \label {eq:limfpn} \lim _{n\rightarrow \infty } f(p_n) = q \end {equation} for every sequence \(\{p_n\}^{\infty }_{n = 1}\) in \(E\) such that \begin {equation} \label {eq:pnp} p_n \neq p \quad \text {for all } n \qquad \text {and}\qquad \lim _{n\rightarrow \infty } p_n = p. \end {equation}
Proof. Suppose (2.1) holds, and choose a sequence \(\{p_n\}\) in \(E\) satisfying (2.3). Let \(\varepsilon > 0\) be given. There is a \(\delta > 0\) such that \(|f(x) - q| < \varepsilon \) whenever \(x \in E\) and \(0 < |x - p| < \delta \). By (2.3) there is an \(N \in \mathbb {N}\) such that \(0 <|p_n - p| < \delta \) for all \(n > N\). Thus \(|f(p_n) - q| < \varepsilon \) for all \(n > N\), which shows that (2.2) holds.
Conversely, suppose (2.1) is false. Then there is some \(\varepsilon > 0\) such that for every \(\delta > 0\) there is a point \(x \in E\) (depending on \(\delta \)) with \(0 < |x - p| < \delta \) but \(|f(x) - q|\geq \varepsilon \). Taking \(\delta = \frac {1}{n}\) for \(n = 1, 2, \dots \), we obtain points \(p_n\in E\) with \(0<|p_n - p|<\frac {1}{n}\) and \(|f(p_n)-q|\geq \varepsilon \). This sequence satisfies (2.3), but (2.2) fails for it. Hence, if (2.2) holds for every sequence satisfying (2.3), then (2.1) holds. □
Proof. Since \(p\) is a limit point of \(E\), there is a sequence satisfying (2.3) (Theorem 1.2.13). If \(q\) and \(q'\) were both limits of \(f\) at \(p\), then by the theorem \(f(p_n)\) would converge to both \(q\) and \(q'\). A sequence has at most one limit, so \(q = q'\). □
Theorem 2.1.4. If \(\lim _{x\rightarrow p} f(x) = q \neq 0\), then there exist \(K > 0\) and \(\delta > 0\) such that \(|f(x)| > K\) for every \(x\in E\) with \(0<|x - p| < \delta \).
Proof. Take \(\varepsilon = \frac {1}{2}|q|>0\) in the definition of the limit: there exists \(\delta > 0\) such that \(|f(x) - q| < \frac {1}{2}|q|\) whenever \(x\in E\) and \(0 < |x - p|< \delta \). For such \(x\), \[|q| - |f(x)| \leq |f(x) - q| < \tfrac {1}{2}|q|, \qquad \text {so}\qquad |f(x)| > \tfrac {1}{2}|q| .\] Hence \(K = \frac {1}{2}|q|\) will do. □
Theorem 2.1.5. Let \(E\subseteq \mathbb {R}\), let \(p\) be a limit point of \(E\), and let \(f\) and \(g\) be real valued functions on \(E\) with \(\lim _{x\rightarrow p} f(x) = A\) and \(\lim _{x\rightarrow p} g(x) = B\). Then
- (a)
- \(\lim _{x\rightarrow p} (f + g)(x) = A + B\);
- (b)
- \(\lim _{x\rightarrow p} (fg)(x) = AB\);
- (c)
- \(\lim _{x\rightarrow p} \left (\frac {f}{g}\right )(x) = \frac {A}{B}\) if \(B\neq 0\).
Proof. Let \(\{p_n\}\) be any sequence in \(E\) satisfying (2.3). By the theorem above, \(f(p_n)\longrightarrow A\) and \(g(p_n)\longrightarrow B\). By the corresponding rules for limits of sequences, \(f(p_n)+g(p_n)\longrightarrow A+B\) and \(f(p_n)g(p_n)\longrightarrow AB\). If \(B\neq 0\), then by the previous theorem \(g(x)\neq 0\) for \(x\) near \(p\), so \(\frac {f}{g}\) is defined there and \(\frac {f(p_n)}{g(p_n)}\longrightarrow \frac {A}{B}\). Since the sequence was arbitrary, the theorem above gives (a), (b) and (c). □
Solution. We have \(|x^2 - 9| = |x-3|\,|x+3|\). If \(|x-3|<1\) then \(2<x<4\), so \(|x+3|<7\) and \(|x^2-9|<7|x-3|\). Given \(\varepsilon >0\), put \(\delta = \min \left \{1,\frac {\varepsilon }{7}\right \}\). Then \(0<|x-3|<\delta \) gives \(|x^2 - 9|<7\cdot \frac {\varepsilon }{7} = \varepsilon \). For example, \(\varepsilon = 0.01\) is achieved with \(\delta = \frac {1}{700}\).
The restriction \(\delta \leq 1\) is a standard device: it keeps \(x\) in a bounded interval so that the extra factor \(|x+3|\) can be bounded by a constant.
Example 2.1.7. Find \(\displaystyle \lim _{x\rightarrow 2}\frac {x^2-4}{x-2}\) and \(\displaystyle \lim _{x\rightarrow 0} x\sin \frac {1}{x}\), justifying each with an explicit \(\delta \).
Solution. For \(x\neq 2\), \(\frac {x^2-4}{x-2} = x+2\), so \(\left |\frac {x^2-4}{x-2} - 4\right | = |x-2|\), and \(\delta = \varepsilon \) works: the limit is \(4\). The function is not defined at \(2\), which does not matter since the definition only uses \(0<|x-2|<\delta \).
For \(x\neq 0\), \(\left |x\sin \frac {1}{x} - 0\right |\leq |x|\), so again \(\delta = \varepsilon \) works and the limit is \(0\).
Solution. Use the sequential form of the limit (the first theorem of this section). The sequences \(p_n = \frac {1}{n\pi }\) and \(q_n = \frac {1}{2n\pi +\frac {\pi }{2}}\) both tend to \(0\) through non-zero values, but \[\sin \frac {1}{p_n} = \sin n\pi = 0 \longrightarrow 0, \qquad \sin \frac {1}{q_n} = \sin \left (2n\pi +\frac {\pi }{2}\right ) = 1\longrightarrow 1 .\] If the limit existed it would equal both \(0\) and \(1\). Hence there is no limit.
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