In this chapter we study limits and continuity of functions \(f: D \to \mathbb{R}\) with domains \(D \subset \mathbb{R}\).
Definition. Let \(D \subset \mathbb{R}\) be a subset of the real numbers. A point \(x \in \mathbb{R}\) is called a limit point of the set D if for every \(\epsilon > 0\)
\(U_\epsilon(x) \cap (D \setminus \{x\}) \neq \emptyset\).
The set of limit points of a set D is denoted by D'.
Remark. It is not difficult to show that a point \(x \in \mathbb{R}\) is a limit point of the set D if and only if there exists a sequence \((x_n)_{n=1}^\infty\) with \(x_n \in D \setminus \{x\}\) for all \(n \in \mathbb{N}\) such that
\(\lim_{n \to \infty} x_n = x\).
This explains the name "limit point".
Examples. i) For real numbers \(a \le b\) we have
\((a, b)' = [a, b]\), \([a, b]' = [a, b]\),
\([a, b)' = (a, b]' = [a, b]\).
ii) For \(D = [a, b] \cup \{c\}\) where \(a < b < c\) we have \(D'=[a, b]\).
iii) We have \(\mathbb{Q}' = \mathbb{R}\).
Definition. Let \(f: D \to \mathbb{R} (D \subset \mathbb{R})\) be a function, \(\xi \in D'\), and let \(y \in \mathbb{R}\). We say that f tends to the limit y as x tends to \(\xi\) if
\(\forall \epsilon > 0 \exists \delta > 0 \forall x \in U_\delta(\xi) \cap D \setminus \{\xi\}: |f(x) - y| < \epsilon\).
In this case we write
\(\lim_{x \to \xi} f(x) = y\) or \(f(x) \to y\) as \(x \to \xi\).
Remark. i) In the definition it is crucial to consider the "deleted" neighbourhood of \(\xi\) (i.e. we have to exclude the point \(\xi\)). To illustrate this, let us look at the function
\(f: \mathbb{R} \to \mathbb{R}, f(x) := \begin{cases} 1, & x \neq 0 \\ 0, & x = 0 \end{cases}\)
Then, according to our definition,
\(\lim_{x \to 0} f(x) = 1\), but \(f(0) \neq 1\).
The limit \(\lim_{x \to 0} f(x)\) would not exist if we did not exclude \(\xi\) in the above definition.
ii) Similar to the limits of sequences, we can show that if it exists, the limit of a function is unique.
Theorem 29.
i) Let \(f: D \to \mathbb{R} (D \subset \mathbb{R})\) be a function, \(\xi \in D'\) and \(y \in \mathbb{R}\). Then we have
\(\lim_{x \to \xi} f(x) = y \iff\) For every sequence \((x_n)_{n=1}^\infty\) with \(x_n \in D \setminus \{\xi\}\) and \(x_n \to \xi\) we have \(\lim_{n \to \infty} f(x_n) = y\).
ii) Let \(f, g: D \to \mathbb{R} (D \subset \mathbb{R})\) be two functions, \(\xi \in D'\), and suppose that \(\lim_{x \to \xi} f(x)\) and \(\lim_{x \to \xi} g(x)\) exist. Then, for any \(\alpha, \beta \in \mathbb{R}\), we have
\(\lim_{x \to \xi} (\alpha f(x) + \beta g(x)) = \alpha \lim_{x \to \xi} f(x) + \beta \lim_{x \to \xi} g(x)\),
\(\lim_{x \to \xi} (f(x) g(x)) = \lim_{x \to \xi} f(x) \cdot \lim_{x \to \xi} g(x)\),
and if \(\lim_{x \to \xi} g(x) \neq 0\) we have
\(\lim_{x \to \xi} \frac{f(x)}{g(x)} = \frac{\lim_{x \to \xi} f(x)}{\lim_{x \to \xi} g(x)}\).
Proof. i) "\(\implies\)" We suppose \(\lim_{x \to \xi} f(x) = y\) and choose a sequence \((x_n)_{n=1}^\infty\) with \(x_n \in D \setminus \{\xi\}\) and \(x_n \to \xi\). We show \(\lim_{n \to \infty} f(x_n) = y\).
Let \(\epsilon > 0\), then we know:
\(\exists \delta > 0 \forall x \in D \setminus \{\xi\} \cap U_\delta(\xi): |f(x) - y| < \epsilon\).
Moreover, we know: \(\exists N \in \mathbb{N} \quad \forall n \ge N: x_n \in U_\delta(\xi)\)
\(\implies \forall n \ge N: |f(x_n) - y| < \epsilon\).
"\(\impliedby\)" We suppose \(\lim_{n \to \infty} f(x_n) = y\) for every \((x_n)_{n=1}^\infty\) with \(x_n \in D \setminus \{\xi\}\) and \(x_n \to \xi\).
We show \(\lim_{x \to \xi} f(x) = y\), i.e.,
\(\forall \epsilon > 0 \exists \delta > 0 \forall x \in D \setminus \{\xi\} \cap U_\delta(\xi): |f(x) - y| < \epsilon\).
We show this by contradiction:
Assume: \(\exists \epsilon > 0 \forall \delta > 0 \exists x \in D \setminus \{\xi\} \cap U_\delta(\xi): |f(x) - y| \ge \epsilon\).
\(\implies \forall n = 1, 2, 3, \dots\) We set \(\delta = \delta_n = \frac{1}{n}\), so that we obtain \(x_n \in D \setminus \{\xi\} \cap U_{1/n}(\xi)\) with \(|f(x_n) - y| \ge \epsilon\).
Hence, we have found a sequence \((x_n)_{n=1}^\infty\) with \(x_n \in D \setminus \{\xi\}\) and \(x_n \to \xi\) such that \(f(x_n) \not\to y\), which is a contradiction.
ii) Follows immediately from part i) and Thm 3.
Examples. i) Let \(p: \mathbb{R} \to \mathbb{R}, p(x) := \sum_{k=0}^n a_k x^k\), be a polynomial of degree \(n \in \mathbb{N}\) and \(\xi \in \mathbb{R}\).
We show \(\lim_{x \to \xi} p(x) = p(\xi)\).
For any \(x \in U_1(\xi)\) we have
\(|p(x) - p(\xi)| = |\sum_{k=0}^n a_k x^k - \sum_{k=0}^n a_k \xi^k| \le \sum_{k=1}^n |a_k| |x^k - \xi^k|\)
\(= \sum_{k=1}^n |a_k| |x-\xi| |\sum_{\nu=0}^{k-1} x^\nu \xi^{k-1-\nu}| \le |x-\xi| \sum_{k=1}^n |a_k| \sum_{\nu=0}^{k-1} |x|^\nu |\xi|^{k-1-\nu}\)
\(\le |x-\xi| \underbrace{\sum_{k=1}^n |a_k| \sum_{\nu=0}^{k-1} (1+|\xi|)^\nu |\xi|^{k-1-\nu}}_{=: M} = M |x-\xi|\),
where we used \(|x| \le 1+|\xi|\).
Hence, for any given \(\epsilon > 0\), setting \(\delta := \min(1, \frac{\epsilon}{M})\), we have for \(x \in U_\delta(\xi): |p(x) - p(\xi)| < \epsilon\).
ii) Let \(f: [0, 1) \cup (1, \infty) \to \mathbb{R}, f(x) := \frac{x-1}{\sqrt{x}-1}, \xi := 1\).
We show \(\lim_{x \to 1} f(x) = 2\).
For any \(x \in U_1(1) \setminus \{1\}\) we have
\(|f(x) - 2| = |\frac{x-1}{\sqrt{x}-1} - 2| = |\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}-1} - 2| = |\sqrt{x}+1 - 2| = |\sqrt{x}-1|\)
\(= |\frac{x-1}{\sqrt{x}+1}| \le |x-1|\).
Hence, for any \(\epsilon > 0\), setting \(\delta := \min(\epsilon, 1)\), we have for \(x \in U_\delta(1) \setminus \{1\}: |f(x) - 2| < \epsilon\).
iii) An analytical definition of sine and cosine is the following:
\(\sin x := \sum_{M=0}^\infty (-1)^M \frac{x^{2M+1}}{(2M+1)!}, \cos x := \sum_{M=0}^\infty (-1)^M \frac{x^{2M}}{(2M)!}\),
for \(x \in \mathbb{R}\).
These series representations often are convenient to compute limits, e.g.
\(\lim_{x \to 0} \frac{\sin x}{x}, \lim_{x \to 0} \frac{1-\cos x}{x^2}, \dots\)
We show \(\lim_{x \to 0} \frac{\sin x}{x} = 1\).
For any \(x \in U_1(0) \setminus \{0\}\) we have
\(|\frac{\sin x}{x} - 1| = |\frac{1}{x} (\sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!}) - 1| = |\sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n+1)!} - 1|\)
\(= |\sum_{n=1}^\infty (-1)^n \frac{x^{2n}}{(2n+1)!}| \le \sum_{n=1}^\infty \frac{|x|^{2n}}{(2n+1)!} = x^2 \sum_{n=1}^\infty \frac{|x|^{2n-2}}{(2n+1)!}\)
\(\le x^2 \underbrace{\sum_{n=1}^\infty \frac{1}{(2n+1)!}}_{=: M} = M x^2 \le M |x|\),
where we used \(|x| < 1\).
Hence, for any given \(\epsilon > 0\), setting \(\delta := \min(1, \frac{\epsilon}{M})\), we have for \(x \in U_\delta(0) \setminus \{0\}\)
\(|\frac{\sin x}{x} - 1| \le M|x| < \epsilon\).
Next we consider one-sided limits.
Definition. Let \(f: D \to \mathbb{R} (D \subset \mathbb{R})\) be a function and \(\xi \in \mathbb{R}\) such that \(\xi \in (D \cap (-\infty, \xi))'\) (i.e. \(\exists\) sequence \((x_n)_{n=1}^\infty\) with \(x_n \in D \cap (-\infty, \xi)\) such that \(x_n \to \xi\)).
We say that f tends to y as x tends to \(\xi\) from the left if
\(\forall \epsilon > 0 \exists \delta > 0 \forall x \in D \cap (\xi-\delta, \xi): |f(x) - y| < \epsilon\).
We write \(\lim_{x \to \xi-} f(x) = y\) or \(f(x) \to y\) as \(x \to \xi-\), and we call this the left-hand limit of f at \(\xi\).
Similarly, if \(\xi \in (D \cap (\xi, \infty))'\), then we say that f tends to y as x tends to \(\xi\) from the right if
\(\forall \epsilon > 0 \exists \delta > 0 \forall x \in D \cap (\xi, \xi+\delta): |f(x) - y| < \epsilon\).
We write \(\lim_{x \to \xi+} f(x) = y\) or \(f(x) \to y\) as \(x \to \xi+\), and call this the right-hand limit of f at \(\xi\).
Remark. i) For a function \(f: D \to \mathbb{R} (D \subset \mathbb{R})\) and \(\xi \in \mathbb{R}\) such that \(\xi \in (D \cap (-\infty, \xi))' \cap (D \cap (\xi, \infty))'\) it is not difficult to show:
\(\lim_{x \to \xi} f(x) = y \iff \lim_{x \to \xi-} f(x) = \lim_{x \to \xi+} f(x) = y\).
ii) Using the same ideas, we could prove an analogous version of Thm 29 for one-sided limits.
Examples. Let \(f: \mathbb{R} \to \mathbb{R}\) be defined by
\(f(x) = \begin{cases} 0, & x < 0 \\ 1, & 0 \le x < 1 \\ 2, & x=1 \\ 1, & x> 1 \end{cases}\)
Then \(\lim_{x \to 0-} f(x) = 0\), \(\lim_{x \to 0+} f(x) = 1\), and \(\lim_{x \to 0} f(x)\) does not exist.
Moreover, \(\lim_{x \to 1-} f(x) = 1\), \(\lim_{x \to 1+} f(x) = 1\), and \(\lim_{x \to 1} f(x)\) exists and is equal to 1.
Next we introduce limits at infinity.
Definition. Let \(f: D \to \mathbb{R} (D \subset \mathbb{R})\) be a function where D is not bounded above (i.e. \(\exists\) sequence \((x_n)_{n=1}^\infty\) in D such that \(x_n \to \infty\)), and let \(y \in \mathbb{R}\).
We say that f tends to y as x tends to infinity if
\(\forall \epsilon > 0 \exists L > 0 \forall x \in D \cap (L, \infty): |f(x) - y| < \epsilon\).
We write \(\lim_{x \to \infty} f(x) = y\) or \(f(x) \to y\) as \(x \to \infty\).
Accordingly, if \(D \subset \mathbb{R}\) is not bounded below, we say that f tends to y as x tends to negative infinity if
\(\forall \epsilon > 0 \exists L < 0 \forall x \in D \cap (-\infty, L): |f(x) - y| < \epsilon\).
We write \(\lim_{x \to -\infty} f(x) = y\) or \(f(x) \to y\) as \(x \to -\infty\).
Remark. i) The convergence of sequences is the case \(D = \mathbb{N}\) in the above definition.
ii) Using the same ideas, we could prove an analogous version of Thm 29 for limits at infinity.
Example. Let \(f: [0, \infty) \to \mathbb{R}\) be defined by \(f(x) := \sqrt{x+1} - \sqrt{x}\). Then we show \(\lim_{x \to \infty} f(x) = 0\).
For \(x > 0\) we have
\(|f(x) - 0| = |\sqrt{x+1} - \sqrt{x}| = \sqrt{x+1} - \sqrt{x}\)
\(= \frac{(\sqrt{x+1} - \sqrt{x})(\sqrt{x+1} + \sqrt{x})}{\sqrt{x+1} + \sqrt{x}} = \frac{x+1 - x}{\sqrt{x+1} + \sqrt{x}} = \frac{1}{\sqrt{x+1} + \sqrt{x}} \le \frac{1}{2\sqrt{x}}\).
Hence, given any \(\epsilon > 0\), we have
\(\frac{1}{2\sqrt{x}} < \epsilon \iff \sqrt{x}> \frac{1}{2\epsilon} \iff x > \frac{1}{4\epsilon^2}\).
We set \(L := \frac{1}{4\epsilon^2}\), then we have for \(x > L\):
\(|f(x) - 0| < \epsilon\).