Limit operations are fundamental because they allow us to compute the limit of a sum, a product, or a quotient starting from the limits of the individual sequences.
In this page we will consider three essential results:
- the limit of a sum;
- the limit of a product;
- the limit of a quotient.
Contents
Limit of a Sum
Let \((a_n)\) and \((b_n)\) be two real sequences such that:
\[ \lim_{n\to\infty}a_n=A \qquad\text{and}\qquad \lim_{n\to\infty}b_n=B. \]
Then:
\[ \lim_{n\to\infty}(a_n+b_n)=A+B. \]
Proof. We want to prove that, for every \(\varepsilon>0\), there exists \(N\in\mathbb{N}\) such that, for every \(n\geq N\),
\[ |(a_n+b_n)-(A+B)|<\varepsilon. \]
We observe that:
\[ |(a_n+b_n)-(A+B)| = |(a_n-A)+(b_n-B)|. \]
By the triangle inequality:
\[ |(a_n-A)+(b_n-B)| \leq |a_n-A|+|b_n-B|. \]
Since \(a_n\to A\), there exists \(N_1\in\mathbb{N}\) such that, for every \(n\geq N_1\),
\[ |a_n-A|<\frac{\varepsilon}{2}. \]
Since \(b_n\to B\), there exists \(N_2\in\mathbb{N}\) such that, for every \(n\geq N_2\),
\[ |b_n-B|<\frac{\varepsilon}{2}. \]
Choose:
\[ N=\max\{N_1,N_2\}. \]
Then, for every \(n\geq N\), we have:
\[ |(a_n+b_n)-(A+B)| \leq |a_n-A|+|b_n-B| < \frac{\varepsilon}{2}+\frac{\varepsilon}{2} = \varepsilon. \]
Therefore:
\[ \lim_{n\to\infty}(a_n+b_n)=A+B. \]
Limit of a Product
Let \((a_n)\) and \((b_n)\) be two real sequences such that:
\[ \lim_{n\to\infty}a_n=A \qquad\text{and}\qquad \lim_{n\to\infty}b_n=B. \]
Then:
\[ \lim_{n\to\infty}(a_n b_n)=AB. \]
Proof. We want to prove 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 write:
\[ a_n b_n-AB = a_n b_n-A b_n+A b_n-AB. \]
Therefore:
\[ a_n b_n-AB = (a_n-A)b_n+A(b_n-B). \]
By the triangle inequality:
\[ |a_n b_n-AB| \leq |a_n-A|\,|b_n|+|A|\,|b_n-B|. \]
Since \(b_n\to B\), the sequence \((b_n)\) is eventually bounded. In particular, choosing \(1>0\), there exists \(N_0\in\mathbb{N}\) such that, for every \(n\geq N_0\),
\[ |b_n-B|<1. \]
Hence:
\[ |b_n| = |b_n-B+B| \leq |b_n-B|+|B| < |B|+1. \]
Let now \(\varepsilon>0\). 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 \(n\geq N_2\),
\[ |b_n-B|<\frac{\varepsilon}{2(|A|+1)}. \]
Choose:
\[ N=\max\{N_0,N_1,N_2\}. \]
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:
\[ |A|\,|b_n-B| \leq (|A|+1)|b_n-B| < (|A|+1)\frac{\varepsilon}{2(|A|+1)} = \frac{\varepsilon}{2}. \]
Thus:
\[ |a_n b_n-AB| \leq |a_n-A|\,|b_n|+|A|\,|b_n-B| < \frac{\varepsilon}{2}+\frac{\varepsilon}{2} = \varepsilon. \]
Therefore:
\[ \lim_{n\to\infty}(a_n b_n)=AB. \]
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 \(B\neq 0\). Assume also that \(b_n\neq 0\) eventually, that is, for all sufficiently large \(n\).
Then:
\[ \lim_{n\to\infty}\frac{a_n}{b_n} = \frac{A}{B}. \]
Proof. Since \(b_n\to B\) and \(B\neq 0\), we can apply the sign preservation theorem. In particular, there exists \(N_0\in\mathbb{N}\) such that, for every \(n\geq N_0\),
\[ |b_n-B|<\frac{|B|}{2}. \]
From this inequality it follows that:
\[ |b_n| \geq |B|-|b_n-B| > |B|-\frac{|B|}{2} = \frac{|B|}{2}. \]
Therefore, for every \(n\geq N_0\),
\[ |b_n|>\frac{|B|}{2}. \]
In particular, \(b_n\neq 0\) eventually.
We now want to estimate:
\[ \left| \frac{a_n}{b_n}-\frac{A}{B} \right|. \]
Rewrite the difference:
\[ \left| \frac{a_n}{b_n}-\frac{A}{B} \right| = \left| \frac{B a_n-A b_n}{B b_n} \right|. \]
Add and subtract \(AB\) in the numerator:
\[ B a_n-A b_n = B a_n-AB+AB-A b_n. \]
Hence:
\[ B a_n-A b_n = B(a_n-A)+A(B-b_n). \]
Therefore, by the triangle inequality:
\[ |B a_n-A b_n| \leq |B|\,|a_n-A|+|A|\,|b_n-B|. \]
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 \(n\geq N_0\), since \(|b_n|>|B|/2\), we obtain:
\[ |B|\,|b_n| > |B|\cdot\frac{|B|}{2} = \frac{|B|^2}{2}. \]
Thus:
\[ \left| \frac{a_n}{b_n}-\frac{A}{B} \right| \leq \frac{2}{|B|^2} \left( |B|\,|a_n-A|+|A|\,|b_n-B| \right). \]
Let now \(\varepsilon>0\). 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 \(n\geq N_2\),
\[ |b_n-B| < \frac{\varepsilon |B|^2}{4(|A|+1)}. \]
Choose:
\[ N=\max\{N_0,N_1,N_2\}. \]
Then, for every \(n\geq N\), we have:
\[ \left| \frac{a_n}{b_n}-\frac{A}{B} \right| \leq \frac{2}{|B|^2} \left( |B|\,|a_n-A|+|A|\,|b_n-B| \right). \]
Using the previous estimates:
\[ |B|\,|a_n-A| < |B|\cdot\frac{\varepsilon |B|}{4} = \frac{\varepsilon |B|^2}{4}. \]
Moreover:
\[ |A|\,|b_n-B| \leq (|A|+1)|b_n-B| < (|A|+1)\frac{\varepsilon |B|^2}{4(|A|+1)} = \frac{\varepsilon |B|^2}{4}. \]
Therefore:
\[ |B|\,|a_n-A|+|A|\,|b_n-B| < \frac{\varepsilon |B|^2}{4} + \frac{\varepsilon |B|^2}{4} = \frac{\varepsilon |B|^2}{2}. \]
Hence:
\[ \left| \frac{a_n}{b_n}-\frac{A}{B} \right| < \frac{2}{|B|^2} \cdot \frac{\varepsilon |B|^2}{2} = \varepsilon. \]
Therefore:
\[ \lim_{n\to\infty}\frac{a_n}{b_n} = \frac{A}{B}. \]