The operations on limits of sequences allow us to compute the limit of a sequence obtained by combining two simpler sequences through their sum, difference, product, or quotient.
The underlying idea is the following: if two sequences \((a_n)\) and \((b_n)\) have finite limits, then, under suitable hypotheses, the sequences obtained from the algebraic operations between \(a_n\) and \(b_n\) also have a limit, and this limit is computed by performing the corresponding operation on the limits themselves.
In this article we 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 the 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 \(+\infty-\infty\), \(0\cdot\infty\), \(\frac{0}{0}\), or \(\frac{\infty}{\infty}\). In such cases a separate analysis is required.
Contents
- Operations on limits of sequences
- Limit of a sum
- Limit of a difference
- Limit of a product
- Limit of a quotient
- Remarks on indeterminate forms
- Examples on 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 we also have
\[ \lim_{n\to+\infty}\frac{a_n}{b_n}=\frac{A}{B}. \]
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 sections that follow give rigorous proofs of these properties using the definition of the limit of a sequence.
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,
\[ \lim_{n\to+\infty}(a_n+b_n)=A+B. \]
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
\[ \lim_{n\to+\infty}(a_n-b_n)=A-B. \]
This result follows directly from the limit of a sum, by noting that
\[ a_n-b_n=a_n+(-b_n). \]
Since \(b_n\to B\), we have
\[ -b_n\to -B. \]
Hence, applying the limit of a sum,
\[ \lim_{n\to+\infty}(a_n-b_n) = \lim_{n\to+\infty}\bigl(a_n+(-b_n)\bigr) = A+(-B) = A-B. \]
Equivalently, the result can be proved directly from the definition of limit. Indeed,
\[ |(a_n-b_n)-(A-B)| = |(a_n-A)-(b_n-B)| \leq |a_n-A|+|b_n-B|. \]
The conclusion then follows exactly as in the case of the sum.
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 \(n\geq N\),
\[ |a_n b_n-AB|<\varepsilon. \]
We rewrite 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 a fundamental fact: every convergent sequence is bounded. Since \(b_n\to B\), there exists a positive real constant \(C\) such that
\[ |b_n|\leq C \]
for all sufficiently large \(n\).
To be explicit, taking \(1>0\), the convergence \(b_n\to B\) yields an \(N_0\in\mathbb{N}\) such that, for every \(n\geq N_0\),
\[ |b_n-B|<1. \]
From this,
\[ |b_n| = |b_n-B+B| \leq |b_n-B|+|B| < |B|+1. \]
Therefore, 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
\[ |a_n-A|\,|b_n| < \frac{\varepsilon}{2(|B|+1)}(|B|+1) = \frac{\varepsilon}{2}. \]
Moreover
\[ |A|\,|b_n-B| \leq (|A|+1)|b_n-B| < (|A|+1)\frac{\varepsilon}{2(|A|+1)} = \frac{\varepsilon}{2}. \]
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,
\[ \lim_{n\to+\infty}(a_n b_n)=AB. \]
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 \(B\neq0\), we may choose the positive distance
\[ \frac{|B|}{2}>0. \]
From the convergence of \((b_n)\) to \(B\), there exists \(N_0\in\mathbb{N}\) such that, for every \(n\geq N_0\),
\[ |b_n-B|<\frac{|B|}{2}. \]
From the triangle inequality it follows that
\[ |B| = |B-b_n+b_n| \leq |B-b_n|+|b_n|. \]
Hence
\[ |b_n| \geq |B|-|B-b_n| = |B|-|b_n-B|. \]
Therefore, 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.
We now estimate the difference between the quotient and the expected limit:
\[ \left|\frac{a_n}{b_n}-\frac{A}{B}\right|. \]
Reducing to a common denominator, we obtain
\[ \left|\frac{a_n}{b_n}-\frac{A}{B}\right| = \left|\frac{B a_n-A b_n}{B b_n}\right|. \]
We add and subtract \(AB\) 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,
\[ |B|\,|a_n-A| < |B|\cdot\frac{\varepsilon |B|}{4} = \frac{\varepsilon |B|^2}{4}. \]
Moreover
\[ |A|\,|b_n-B| \leq (|A|+1)|b_n-B| < (|A|+1)\frac{\varepsilon |B|^2}{4(|A|+1)} = \frac{\varepsilon |B|^2}{4}. \]
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 preceding rules were 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 appear, or when denominators tend to zero. In such cases the algebraic rules cannot always be applied directly.
For instance, expressions of the form
\[ +\infty-\infty, \qquad 0\cdot\infty, \qquad \frac{0}{0}, \qquad \frac{+\infty}{+\infty} \]
are called indeterminate forms. The word "indeterminate" means that knowledge of the limits of the individual parts alone is not enough to determine the limit of the whole expression.
For example, if \(a_n\to+\infty\) and \(b_n\to+\infty\), the limit of \(a_n-b_n\) is not automatically determined. 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 in different ways depending on the sequences involved.
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,
\[ \frac{\displaystyle \frac{1}{n}}{\displaystyle \frac{1}{n^2}}=n, \]
so the limit is \(+\infty\).
This shows that the information "numerator tending to \(0\)" and "denominator tending to \(0\)" alone is not enough 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 theorems are satisfied.
Examples on operations with limits
Example 1 (limit of a sum). Consider the sequence
\[ c_n=\frac{1}{n}+\frac{n}{n+1}. \]
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
\[ c_n=\frac{n}{n+1}-\frac{1}{n}. \]
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
\[ c_n= \left(2+\frac{1}{n}\right) \left(3-\frac{1}{n}\right). \]
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
\[ c_n= \frac{2+\displaystyle \frac{1}{n}}{3-\displaystyle \frac{1}{n}}. \]
The numerator tends to \(2\); indeed,
\[ \lim_{n\to+\infty}\left(2+\frac{1}{n}\right)=2. \]
The denominator tends to \(3\); indeed,
\[ \lim_{n\to+\infty}\left(3-\frac{1}{n}\right)=3. \]
Since the limit of the denominator is nonzero, 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 caution about quotients whose denominator tends to zero). Consider the sequence
\[ c_n=\frac{\displaystyle \frac{1}{n}}{\displaystyle \frac{1}{n}}. \]
Both the numerator and the denominator tend to \(0\):
\[ \lim_{n\to+\infty}\frac{1}{n}=0. \]
Nevertheless, we cannot apply the theorem on the limit of a quotient directly, because the limit of the denominator is \(0\).
In this case, simplifying, we obtain
\[ c_n=1 \]
for every \(n\in\mathbb{N}_{\ge 1}\), so
\[ \lim_{n\to+\infty}c_n=1. \]
This example shows that a form of the type \(\displaystyle \frac{0}{0}\) must be studied separately: its limit cannot be determined by applying the quotient rule automatically, because the limit of the denominator equals \(0\).
In conclusion, the operations on limits make it possible to compute many limits of sequences in a simple and rigorous way, provided the hypotheses of the theorems are respected. In particular, for the limit of a quotient it is essential that the limit of the denominator be nonzero.