Properties of limits

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We see some basic properties of the limit of a sequence, such as

  • The limit of a sequence is unique.
  • A convergent sequence is bounded.
  • The squeeze theorem.
  • Limiting operation is linear.
  • The limit of the product of sequences is the product of the limits.
  • etc.


Theorem (Uniqueness of limit)

If the sequence \(\{a_n\}\) converges, its limit is unique. In other words, if we have \(\lim_{n\to\infty}a_n = \alpha\) and \(\lim_{n\to\infty}a_n = \beta\), then \(\alpha = \beta\).

Proof. Suppose that \(\{a_n\}\) converges to \(\alpha\) as well as to \(\beta\). Take any positive real number \(\varepsilon\). 

Since \(\lim_{n\to\infty}a_n = \alpha\), there exists \(N'\in\mathbb{N}\) such that for any \(n\geq N'\), \(|a_n - \alpha| < \varepsilon\).

Similarly, there exists \(N'' \in\mathbb{N}\) such that for any \(n\geq N''\), \(|a_n - \beta| < \varepsilon\).

Let \(N = \max\{N', N''\}\). Then, if \(n \geq N\),

\[|\alpha - \beta| = |\alpha - a_n + a_n -\beta| \leq |\alpha - a_n| + |a_n - \beta| < \varepsilon + \varepsilon = 2\varepsilon.\]

Since \(\varepsilon > 0\) is arbitrary, it must be that \(\alpha = \beta\). ■

Definition (Bounded sequence)

Given a sequence \(\{a_n\}\), we can define a set of real numbers that appear in the sequence: \(\{a_n\mid n\in\mathbb{N}\} \subset \mathbb{R}\). If this set is bounded above, bounded below, or bounded, we say that the sequence \(\{a_n\}\) is bounded above, bounded below, or bounded, respectively.

Theorem (Convergent sequences are bounded)

Any convergent sequence (i.e., a sequence that converges) is bounded.

Proof. Let \(\{a_n\}\) be a converging sequence such that \(\lim_{n\to\infty}a_n = \alpha\). Let \(\varepsilon = 1\). Then there exists \(N\in\mathbb{N}\) such that for all \(n\geq N\), \(|a_n -\alpha| < 1\). Then, by the triangle inequality,

\[|a_n| - |\alpha| \leq |a_n - \alpha| < 1\]

so that

\[|a_n| < |\alpha| + 1.\]

Now, let us define

\[M = \max\{|\alpha| + 1, |a_1|, |a_2|, \cdots, |a_{N-1}|\}.\]

Then, whether \(n\geq N\) or \(n<N\), we always have \(|a_n| \leq M\). Hence \(\{a_n\}\) is bounded. ■

Remark. The converse of this theorem is not necessarily true. e.g., \(a_n = (-1)^n\) is bounded, but not convergent. □

Theorem (Squeeze Theorem)

Let \(\{a_n\}, \{b_n\}, \{c_n\}\) be sequences. Suppose that for all \(n\in\mathbb{N}\), we have \(a_n \leq b_n \leq c_n\) and \(\lim_{n\to\infty}a_n = \lim_{n\to\infty}c_n = \alpha\). Then \(\lim_{n\to\infty}b_n = \alpha\).

Proof. Let \(\varepsilon > 0\) be an arbitrary positive real number. Since \(a_n \to \alpha\), there exists \(N'\in\mathbb{N}\) such that for any \(n\geq N'\),  \(|a_n - \alpha| < \varepsilon\). In particular, for any \(n \geq N'\),

\[-\varepsilon < a_n - \alpha.\]

Similarly, there exists \(N'' \in\mathbb{N}\) such that for any \(n \geq N''\),

\[c_n - \alpha < \varepsilon.\]

Let \(N = \max\{N', N''\}\). Then for any \(n \geq N\), we have

\[-\varepsilon < a_n -\alpha \leq c_n - \alpha < \varepsilon.\]

However, since \(a_n -\alpha \leq b_n - \alpha \leq c_n - \alpha\), we have

\[|b_n - \alpha| < \varepsilon.\]

Therefore \(\lim_{n\to\infty}b_n = \alpha\). ■

Let \(\{a_n\}\) be a sequence. Let's pick a subset \(\{n_k\}\) of \(\mathbb{N}\) which we arrange in increasing order:

\[n_1 < n_2 < n_3 <\cdots n_k < n_{k+1} < \cdots.\]

Then, for \(k = 1, 2, \cdots\) we obtain a sequence

\[a_{n_1}, a_{n_2}, \cdots, a_{n_k}, a_{n_{k+1}}, \cdots\]

This new sequence \(\{a_{n_k}\}\) is called a subsequence of \(\{a_n\}\).

Theorem (Convergence of subsequence)

Let \(\{a_n\}\) be a sequence that converges to \(\alpha \in \mathbb{R}\). Then, any subsequence \(\{a_{n_k}\}\) also converges to \(\alpha\).

Proof. Let \(\varepsilon > 0\) be any positive real number. Since \(\lim_{n\to\infty}a_n = \alpha\), there exists \(N\in\mathbb{N}\) such that for all \(n \geq N\), \(|a_n - \alpha| < \varepsilon\). 

Now take \(K\in\mathbb{N}\) such that \(n_{K} \geq N\). Then, for any \(k \geq K\), \(n_k \geq N\). Thus \(|a_{n_k} - \alpha| < \varepsilon\). 

That is, for any \(\varepsilon > 0\), there exists \(K\in\mathbb{N}\) such that for all \(k \geq K\), \(|a_{n_k} - \alpha| < \varepsilon\). This means \(\lim_{k\to\infty}a_{n_k} = \alpha\). ■

Theorem

Let \(\{a_n\}\) be a sequence that converges to \(\alpha \in \mathbb{R}\).

  1. If there exists a real number \(a\) such that \(a_n \geq a\) holds for infinitely many \(n\in\mathbb{N}\), then \(\alpha \geq a\).
  2. If there exists a real number \(b\) such that \(a_n \leq b\) holds for infinitely many \(n\in\mathbb{N}\), then \(\alpha \leq b\).
Proof. We only prove part 1. Part 2 is similar.

Let us define a subsequence of \(\{a_n\}\) such that \(\{a_n\mid a_n \geq a\}\), and re-index this subsequence so that we may assume \(a_n\geq a\) for all \(n\in\mathbb{N}\) in this subsequence. We prove it by contradiction.

Suppose \(\alpha < a\). Let \(\varepsilon = a - \alpha (> 0)\). Since \(\lim_{n\to\infty}a_n = \alpha\), there exists \(N\in\mathbb{N}\) such that for all \(n\geq N\), \(|a_n -\alpha| < \varepsilon\). But then, \(a_n < \alpha + \epsilon = a\), which is a contradiction. ■

Theorem

Let \(\{a_n\}\) and \(\{b_n\}\) be sequences such that \(\lim_{n\to\infty}a_n = \alpha\) and \(\lim_{n\to\infty}b_n = \beta\).

  1. \(\lim_{n\to\infty}ka_n = k\alpha\) where \(k\) is a constant.
  2. \(\lim_{n\to\infty}(a_n + b_n) = \alpha + \beta\).
  3. \(\lim_{n\to\infty}a_nb_n = \alpha\beta\).
  4. \(\lim_{n\to\infty}\frac{a_n}{b_n} = \frac{\alpha}{\beta}\) given \(\beta \neq 0\).
Proof. Let \(\varepsilon > 0\). There exists \(N_1\in\mathbb{N}\) such that for any \(n\geq N_1\), \(|a_n - \alpha| < \varepsilon\).

Also, there exists \(N_2\in\mathbb{N}\) such that for any \(n\geq N_2\), \(|b_n - \beta| < \varepsilon\). Let \(N = \max\{N_1, N_2\}\). Then for \(n\geq N\), we have both \(|a_n - \alpha| < \varepsilon\) and \(|b_n - \beta| < \varepsilon\).
  1. This is a special case of Part 3 (where \(b_n = k\)).
  2. By the triangular inequality, \[ \begin{eqnarray} |(a_n + b_n) - (\alpha + \beta)| &=& |(a_n - \alpha) + (b_n - \beta)|\\ &\leq & |a_n - \alpha| + |b_n - \beta|\\ &< & \varepsilon + \varepsilon = 2\varepsilon. \end{eqnarray} \] Note that \(2\varepsilon\) is also an arbitrary positive real number. Therefore, for any \(n\geq N\), we have \[ |(a_n + b_n) - (\alpha + \beta)| < 2\varepsilon.\] Thus, \[\lim_{n\to\infty}(a_n + b_n) = \alpha + \beta. \]
  3. \(\{a_n\}\) and \(\{b_n\}\) are convergent, and are hence bounded. Therefore, there exists some real number \(M>0\) such that \[|a_n| \leq M, ~ |b_n| \leq M.\] By the above theorem, we also have \[|\alpha| \leq M, ~ |\beta| \leq M.\] Then, \[ \begin{eqnarray} |a_nb_n - \alpha\beta| &=& |a_nb_n - a_n\beta + a_n\beta - \alpha\beta|\\ &=& |a_n(b_n - \beta) + (a_n - \alpha)\beta|\\ &\leq&|a_n|\cdot|b_n - \beta| + |a_n - \alpha|\cdot|\beta|\\ & < &M\varepsilon + \varepsilon M = 2M\varepsilon. \end{eqnarray} \] Note that \(2M\varepsilon\) is an arbitrary positive real number. Thus, \[\lim_{n\to\infty}a_nb_n = \alpha\beta.\]
  4. First we show that there exists \(N\in\mathbb{N}\) such that for any \(n\geq N\), \(b_n \neq 0\). Since \(\beta \neq 0\), \(|\beta| > 0\). There exists \(N\) such that for any \(n\geq N\), \(|b_n - \beta| < \frac{1}{2}|\beta|\). By the triangular inequality, \[|\beta| - |b_n| \leq |\beta - b_n| < \frac{1}{2}|\beta|\] so that \[|b_n| > |\beta| - \frac{1}{2}|\beta| = \frac{1}{2}|\beta| > 0.\] Therefore, for \(n\geq N\), \(b_n \neq 0\). For any \(\varepsilon > 0\) and a sufficiently large \(n\), we have \(|b_n - \beta| < \varepsilon\), and \[\left|\frac{1}{b_n} - \frac{1}{\beta}\right| = \frac{|\beta - b_n|}{|b_n\beta|} < \frac{2\varepsilon}{\beta^2}.\] Thus, \(\{\frac{1}{b_n}\}\) converges to \(\frac{1}{\beta}\). Therefore, using Part 3, we have \[\lim_{n\to\infty}\frac{a_n}{b_n} = \lim_{n\to\infty}{a_n}{\frac{1}{b_n}} = \alpha\cdot\frac{1}{\beta} = \frac{\alpha}{\beta}.\]
■

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