The algebra of limits of sequences allows one to compute the limit of a sequence obtained by combining two simpler sequences by addition, subtraction, multiplication or division.
The underlying idea is the following: if two sequences \((a_n)\) and \((b_n)\) have finite limits, then, under suitable hypotheses, the sequences obtained by performing algebraic operations term by term on \(a_n\) and \(b_n\) also have limits, and these limits can be computed by applying the same operations to the individual limits.
Consider the case in which
\[ \lim_{n\to+\infty}a_n=A \qquad\text{and}\qquad \lim_{n\to+\infty}b_n=B, \]
with \(A,B\in\mathbb{R}\). We shall therefore study operations on finite limits of convergent real sequences.
It should be stressed from the outset that the algebraic rules for limits cannot be applied automatically in the presence of indeterminate forms, such as \(+\infty-\infty\), \(0\cdot\infty\), \(\displaystyle \frac{0}{0}\) or \(\displaystyle \frac{\infty}{\infty}\). In such cases a specific analysis is required.
Contents
- Operations on limits of sequences
- The limit of a sum
- The limit of a difference
- The limit of a product
- The limit of a quotient
- Remarks on indeterminate forms
- Examples of operations with limits
Operations on limits of sequences
Let \((a_n)\) and \((b_n)\) be two convergent real sequences, that is, such that
\[ \lim_{n\to+\infty}a_n=A, \qquad \lim_{n\to+\infty}b_n=B, \]
with \(A,B\in\mathbb{R}\).
Under these hypotheses the following rules hold:
\[ \lim_{n\to+\infty}(a_n+b_n)=A+B, \]
\[ \lim_{n\to+\infty}(a_n-b_n)=A-B, \]
\[ \lim_{n\to+\infty}(a_n b_n)=AB. \]
Moreover, if \(B\neq0\), then \(b_n\neq0\) eventually, and one also has
\[ \lim_{n\to+\infty}\frac{a_n}{b_n}=\frac{A}{B}. \]
The condition \(B\neq0\) in the limit of the quotient is essential. Indeed, if the limit of the denominator were \(0\), one could not conclude in general that the quotient has a finite limit.
The following sections prove these properties rigorously, starting from the definition of the limit of a sequence.
The limit of a sum
Let \((a_n)\) and \((b_n)\) be two real sequences such that
\[ \lim_{n\to+\infty}a_n=A \qquad\text{and}\qquad \lim_{n\to+\infty}b_n=B, \]
with \(A,B\in\mathbb{R}\). Then
\[ \lim_{n\to+\infty}(a_n+b_n)=A+B. \]
Proof. We wish to show that, for every \(\varepsilon>0\), there exists \(N\in\mathbb{N}\) such that, for every \(n\geq N\),
\[ |(a_n+b_n)-(A+B)|<\varepsilon. \]
Observe that
\[ (a_n+b_n)-(A+B)=(a_n-A)+(b_n-B). \]
Applying the triangle inequality, we obtain
\[ |(a_n+b_n)-(A+B)| = |(a_n-A)+(b_n-B)| \leq |a_n-A|+|b_n-B|. \]
Since \(a_n\to A\), for \(\displaystyle \frac{\varepsilon}{2}>0\) there exists \(N_1\in\mathbb{N}\) such that, for every \(n\geq N_1\),
\[ |a_n-A|<\frac{\varepsilon}{2}. \]
Since \(b_n\to B\), for \(\displaystyle \frac{\varepsilon}{2}>0\) there exists \(N_2\in\mathbb{N}\) such that, for every \(n\geq N_2\),
\[ |b_n-B|<\frac{\varepsilon}{2}. \]
Set
\[ N=\max\{N_1,N_2\}. \]
Then, for every \(n\geq N\), both of the preceding inequalities hold. Consequently
\[ |(a_n+b_n)-(A+B)| \leq |a_n-A|+|b_n-B| < \frac{\varepsilon}{2}+\frac{\varepsilon}{2} = \varepsilon. \]
By the definition of limit,
\[ \lim_{n\to+\infty}(a_n+b_n)=A+B. \]
The limit of a difference
Let \((a_n)\) and \((b_n)\) be two real sequences such that
\[ \lim_{n\to+\infty}a_n=A \qquad\text{and}\qquad \lim_{n\to+\infty}b_n=B, \]
with \(A,B\in\mathbb{R}\). Then
\[ \lim_{n\to+\infty}(a_n-b_n)=A-B. \]
Proof. We wish to show that, for every \(\varepsilon>0\), there exists \(N\in\mathbb{N}\) such that, for every \(n\geq N\),
\[ |(a_n-b_n)-(A-B)|<\varepsilon. \]
Observe that
\[ (a_n-b_n)-(A-B)=(a_n-A)-(b_n-B). \]
Applying the triangle inequality, we obtain
\[ |(a_n-b_n)-(A-B)|=|(a_n-A)-(b_n-B)|\leq |a_n-A|+|b_n-B|. \]
Since \(a_n\to A\), for \(\displaystyle \frac{\varepsilon}{2}>0\) there exists \(N_1\in\mathbb{N}\) such that, for every \(n\geq N_1\),
\[ |a_n-A|<\frac{\varepsilon}{2}. \]
Since \(b_n\to B\), for \(\displaystyle \frac{\varepsilon}{2}>0\) there exists \(N_2\in\mathbb{N}\) such that, for every \(n\geq N_2\),
\[ |b_n-B|<\frac{\varepsilon}{2}. \]
Set
\[ N=\max\{N_1,N_2\}. \]
Then, for every \(n\geq N\),
\[ |(a_n-b_n)-(A-B)|\leq |a_n-A|+|b_n-B|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. \]
By the definition of limit,
\[ \lim_{n\to+\infty}(a_n-b_n)=A-B. \]
The limit of a product
Let \((a_n)\) and \((b_n)\) be two real sequences such that
\[ \lim_{n\to+\infty}a_n=A \qquad\text{and}\qquad \lim_{n\to+\infty}b_n=B, \]
with \(A,B\in\mathbb{R}\). Then
\[ \lim_{n\to+\infty}(a_n b_n)=AB. \]
Proof. We wish to show that, for every \(\varepsilon>0\), there exists \(N\in\mathbb{N}\) such that, for every \(n\geq N\),
\[ |a_n b_n-AB|<\varepsilon. \]
Let us write the difference in a convenient form:
\[ a_n b_n-AB = a_n b_n-A b_n+A b_n-AB. \]
Hence
\[ a_n b_n-AB = (a_n-A)b_n+A(b_n-B). \]
Applying the triangle inequality, we obtain
\[ |a_n b_n-AB| \leq |a_n-A|\,|b_n|+|A|\,|b_n-B|. \]
We now use the fact that a convergent sequence is eventually bounded. Since \(b_n\to B\), applying the definition of limit with \(\varepsilon=1\), there exists \(N_0\in\mathbb{N}\) such that, for every \(n\geq N_0\),
\[ |b_n-B|<1. \]
Hence
\[ |b_n| = |b_n-B+B| \leq |b_n-B|+|B| < |B|+1. \]
Thus, eventually,
\[ |b_n|<|B|+1. \]
Now fix \(\varepsilon>0\). Since \(a_n\to A\), there exists \(N_1\in\mathbb{N}\) such that, for every \(n\geq N_1\),
\[ |a_n-A|<\frac{\varepsilon}{2(|B|+1)}. \]
Since \(b_n\to B\), there exists \(N_2\in\mathbb{N}\) such that, for every \(n\geq N_2\),
\[ |b_n-B|<\frac{\varepsilon}{2(|A|+1)}. \]
Set
\[ N=\max\{N_0,N_1,N_2\}. \]
Then, for every \(n\geq N\), we have
\[ |b_n|<|B|+1, \qquad |a_n-A|<\frac{\varepsilon}{2(|B|+1)} \]
and
\[ |b_n-B|<\frac{\varepsilon}{2(|A|+1)}. \]
Therefore
\[ |a_n-A|\,|b_n| < \frac{\varepsilon}{2(|B|+1)}(|B|+1) = \frac{\varepsilon}{2}. \]
Furthermore
\[ |A|\,|b_n-B| \leq |A|\frac{\varepsilon}{2(|A|+1)} = \frac{|A|}{|A|+1}\cdot\frac{\varepsilon}{2} < \frac{\varepsilon}{2}. \]
Consequently
\[ |a_n b_n-AB| \leq |a_n-A|\,|b_n|+|A|\,|b_n-B| < \frac{\varepsilon}{2}+\frac{\varepsilon}{2} = \varepsilon. \]
By the definition of limit,
\[ \lim_{n\to+\infty}(a_n b_n)=AB. \]
The limit of a quotient
Let \((a_n)\) and \((b_n)\) be two real sequences such that
\[ \lim_{n\to+\infty}a_n=A \qquad\text{and}\qquad \lim_{n\to+\infty}b_n=B, \]
with \(A,B\in\mathbb{R}\) and \(B\neq0\). Then \(b_n\neq0\) eventually, and
\[ \lim_{n\to+\infty}\frac{a_n}{b_n}=\frac{A}{B}. \]
Proof. Since \(b_n\to B\) and \(B\neq0\), the number
\[ \frac{|B|}{2} \]
is positive.
By the convergence of \((b_n)\) to \(B\), there exists \(N_0\in\mathbb{N}\) such that, for every \(n\geq N_0\),
\[ |b_n-B|<\frac{|B|}{2}. \]
By the reverse triangle inequality it follows that
\[ |b_n|=|B+(b_n-B)|\geq |B|-|b_n-B|. \]
Hence, for every \(n\geq N_0\),
\[ |b_n|>|B|-\frac{|B|}{2}=\frac{|B|}{2}. \]
In particular, \(b_n\neq0\) for every \(n\geq N_0\). This shows that the quotient \(\frac{a_n}{b_n}\) is well defined eventually.
Let us now estimate the difference between the quotient and the expected limit:
\[ \left|\frac{a_n}{b_n}-\frac{A}{B}\right|. \]
Bringing the expression to a common denominator, we obtain
\[ \left|\frac{a_n}{b_n}-\frac{A}{B}\right| = \left|\frac{B a_n-A b_n}{B b_n}\right|. \]
Adding and subtracting \(AB\) in the numerator,
\[ B a_n-A b_n = B a_n-AB+AB-A b_n. \]
Hence
\[ B a_n-A b_n = B(a_n-A)+A(B-b_n). \]
Applying the triangle inequality,
\[ |B a_n-A b_n| \leq |B|\,|a_n-A|+|A|\,|B-b_n|. \]
Since
\[ |B-b_n|=|b_n-B|, \]
we obtain
\[ |B a_n-A b_n| \leq |B|\,|a_n-A|+|A|\,|b_n-B|. \]
Consequently
\[ \left|\frac{a_n}{b_n}-\frac{A}{B}\right| \leq \frac{|B|\,|a_n-A|+|A|\,|b_n-B|}{|B|\,|b_n|}. \]
For \(n\geq N_0\), we know that
\[ |b_n|>\frac{|B|}{2}. \]
Therefore
\[ |B|\,|b_n| > |B|\cdot\frac{|B|}{2} = \frac{|B|^2}{2}. \]
Hence, for every \(n\geq N_0\),
\[ \left|\frac{a_n}{b_n}-\frac{A}{B}\right| \leq \frac{2}{|B|^2} \left( |B|\,|a_n-A|+|A|\,|b_n-B| \right). \]
Now fix \(\varepsilon>0\). Since \(a_n\to A\), there exists \(N_1\in\mathbb{N}\) such that, for every \(n\geq N_1\),
\[ |a_n-A|<\frac{\varepsilon |B|}{4}. \]
Since \(b_n\to B\), there exists \(N_2\in\mathbb{N}\) such that, for every \(n\geq N_2\),
\[ |b_n-B|<\frac{\varepsilon |B|^2}{4(|A|+1)}. \]
Set
\[ N=\max\{N_0,N_1,N_2\}. \]
Then, for every \(n\geq N\), all of the preceding estimates hold. In particular,
\[ |B|\,|a_n-A| < |B|\cdot\frac{\varepsilon |B|}{4} = \frac{\varepsilon |B|^2}{4}. \]
Furthermore
\[ |A|\,|b_n-B| \leq |A|\frac{\varepsilon |B|^2}{4(|A|+1)} = \frac{|A|}{|A|+1}\cdot\frac{\varepsilon |B|^2}{4} < \frac{\varepsilon |B|^2}{4}. \]
Adding these,
\[ |B|\,|a_n-A|+|A|\,|b_n-B| < \frac{\varepsilon |B|^2}{4} + \frac{\varepsilon |B|^2}{4} = \frac{\varepsilon |B|^2}{2}. \]
Therefore
\[ \left|\frac{a_n}{b_n}-\frac{A}{B}\right| < \frac{2}{|B|^2}\cdot\frac{\varepsilon |B|^2}{2} = \varepsilon. \]
By the definition of limit,
\[ \lim_{n\to+\infty}\frac{a_n}{b_n}=\frac{A}{B}. \]
Remarks on indeterminate forms
The rules established above have been proved in the case where the sequences \((a_n)\) and \((b_n)\) have finite real limits. In this setting the operations behave in the natural way:
\[ a_n\to A,\quad b_n\to B \quad\Longrightarrow\quad a_n+b_n\to A+B, \]
\[ a_n b_n\to AB, \]
and, if \(B\neq0\),
\[ \frac{a_n}{b_n}\to\frac{A}{B}. \]
Care must be taken, however, when infinite limits occur, or when a quotient has a denominator tending to zero. In such cases it is not always possible to apply the algebraic rules directly.
For instance, expressions of the type
\[ +\infty-\infty, \qquad 0\cdot\infty, \qquad \frac{0}{0}, \qquad \frac{+\infty}{+\infty} \]
are called indeterminate forms. The term "indeterminate" means that knowledge of the limits of the individual parts alone does not suffice to determine the limit of the overall expression.
For example, if \(a_n\to+\infty\) and \(b_n\to+\infty\), the limit of \(a_n-b_n\) is not determined automatically. It may be a real number, it may be \(+\infty\), it may be \(-\infty\), or it may fail to exist.
Likewise, if \(a_n\to0\) and \(b_n\to0\), the quotient
\[ \frac{a_n}{b_n} \]
may behave differently depending on the sequences under consideration.
For example,
\[ \frac{\displaystyle \frac{1}{n}}{\displaystyle \frac{1}{n}}=1 \]
for every \(n\in\mathbb{N}_{\ge 1}\), so the limit is \(1\). On the other hand,
\[ \frac{\displaystyle \frac{1}{n}}{\displaystyle \frac{1}{n^2}}=n, \]
and so the limit is \(+\infty\).
This shows that the mere information that "the numerator tends to \(0\)" and "the denominator tends to \(0\)" does not suffice to determine the limit of the quotient.
For this reason, the rules for operations with limits must be applied only when the hypotheses of the corresponding theorems are satisfied.
Examples of operations with limits
Example 1 (limit of a sum). Consider the sequence
\[ c_n=\frac{1}{n}+\frac{n}{n+1}, \qquad n\in\mathbb{N}_{\ge 1}. \]
We know that
\[ \lim_{n\to+\infty}\frac{1}{n}=0 \]
and
\[ \lim_{n\to+\infty}\frac{n}{n+1}=1. \]
By the limit of a sum,
\[ \lim_{n\to+\infty} \left( \frac{1}{n}+\frac{n}{n+1} \right) = 0+1 = 1. \]
Example 2 (limit of a difference). Consider the sequence
\[ c_n=\frac{n}{n+1}-\frac{1}{n}, \qquad n\in\mathbb{N}_{\ge 1}. \]
Since
\[ \lim_{n\to+\infty}\frac{n}{n+1}=1 \]
and
\[ \lim_{n\to+\infty}\frac{1}{n}=0, \]
by the limit of a difference we obtain
\[ \lim_{n\to+\infty} \left( \frac{n}{n+1}-\frac{1}{n} \right) = 1-0 = 1. \]
Example 3 (limit of a product). Consider the sequence
\[ c_n= \left(2+\frac{1}{n}\right) \left(3-\frac{1}{n}\right), \qquad n\in\mathbb{N}_{\ge 1}. \]
We have
\[ \lim_{n\to+\infty}\left(2+\frac{1}{n}\right)=2 \]
and
\[ \lim_{n\to+\infty}\left(3-\frac{1}{n}\right)=3. \]
By the limit of a product,
\[ \lim_{n\to+\infty} \left(2+\frac{1}{n}\right) \left(3-\frac{1}{n}\right) = 2\cdot3 = 6. \]
Example 4 (limit of a quotient). Consider the sequence
\[ c_n= \frac{2+\displaystyle \frac{1}{n}}{3-\displaystyle \frac{1}{n}}, \qquad n\in\mathbb{N}_{\ge 1}. \]
The numerator tends to \(2\), since
\[ \lim_{n\to+\infty}\left(2+\frac{1}{n}\right)=2. \]
The denominator tends to \(3\), since
\[ \lim_{n\to+\infty}\left(3-\frac{1}{n}\right)=3. \]
Since the limit of the denominator is different from zero, we may apply the limit of a quotient:
\[ \lim_{n\to+\infty} \frac{2+\displaystyle \frac{1}{n}}{3-\displaystyle \frac{1}{n}} = \frac{2}{3}. \]
Example 5 (a caveat: quotient with denominator tending to zero). Consider the sequence
\[ c_n=\frac{\displaystyle \frac{1}{n}}{\displaystyle \frac{1}{n}}, \qquad n\in\mathbb{N}_{\ge 1}. \]
The numerator and the denominator both tend to \(0\), since
\[ \lim_{n\to+\infty}\frac{1}{n}=0. \]
Nevertheless, we cannot apply the theorem on the limit of a quotient directly, because the limit of the denominator is \(0\).
In this case, upon simplification, we obtain
\[ c_n=1 \]
for every \(n\in\mathbb{N}_{\ge 1}\), and hence
\[ \lim_{n\to+\infty}c_n=1. \]
This example shows that a form of the type \(\displaystyle \frac{0}{0}\) must be studied separately: its limit cannot be determined by automatically applying the quotient rule, because the limit of the denominator is equal to \(0\).
In conclusion, operations on limits allow one to compute many limits of sequences in a simple and rigorous way, provided that the hypotheses of the theorems are satisfied. In particular, for the limit of a quotient it is essential that the limit of the denominator be different from zero.