Limit operations are of fundamental importance because they allow us to calculate the limit of a sum, of a product, or of a quotient by deducing it directly from the limits of the individual sequences. These rules considerably simplify calculations, allowing us to analyze the behavior of complex functions without having to resort to more elaborate methods.
Index
Limit of a Sum
Let \(\{a_n\}\) and \(\{b_n\}\) be two sequences. If:
\[ \lim_{n \to \infty} a_n = A \quad \text{and} \quad \lim_{n \to \infty} b_n = B, \]
then:
\[ \lim_{n \to \infty} (a_n + b_n) = A + B. \]
Proof. To prove this theorem, we use the definition of limit for sequences. According to the definition, \(\lim_{n \to \infty} a_n = A\) means that for every \(\epsilon > 0\), there exists a natural number \(N_1\) such that:
\[ |a_n - A| < \epsilon \quad \text{for all} \ n \geq N_1. \]
Similarly, \(\lim_{n \to \infty} b_n = B\) means that for every \(\epsilon > 0\), there exists a natural number \(N_2\) such that:
\[ |b_n - B| < \epsilon \quad \text{for all} \ n \geq N_2. \]
Now, we must prove that:
\[ \lim_{n \to \infty} (a_n + b_n) = A + B. \]
To prove that \(\lim_{n \to \infty} (a_n + b_n) = A + B\), we must show that for every \(\epsilon > 0\), there exists a natural number \(N\) such that:
\[ |(a_n + b_n) - (A + B)| < \epsilon \quad \text{for all} \ n \geq N. \]
We have:
\begin{align} |(a_n + b_n) - (A + B)| &= |(a_n + b_n) - (A + B)| \\ &= |(a_n - A) + (b_n - B)| \end{align}
The triangle inequality ensures that:
\[ |(a_n - A) + (b_n - B)| \leq |a_n - A| + |b_n - B| \]
Now, for every \(\epsilon > 0\), since \(\lim_{n \to \infty} a_n = A\), there exists a natural number \(N_1\) such that:
\[ |a_n - A| < \frac{\epsilon}{2} \quad \text{for all} \ n \geq N_1. \]
Similarly, since \(\lim_{n \to \infty} b_n = B\), there exists a natural number \(N_2\) such that:
\[ |b_n - B| < \frac{\epsilon}{2} \quad \text{for all} \ n \geq N_2. \]
If we choose \( N = \max(N_1, N_2) \), we have that for all \(n \geq N\) the following inequalities hold:
\[ |a_n - A| < \frac{\epsilon}{2} \quad \text{and} \quad |b_n - B| < \frac{\epsilon}{2} \]
Now, for all \(n \geq N\):
\[ |(a_n + b_n) - (A + B)| \leq |a_n - A| + |b_n - B| \]
Since:
\[ |a_n - A| < \frac{\epsilon}{2} \quad \text{and} \quad |b_n - B| < \frac{\epsilon}{2} \]
we have:
\[ |(a_n + b_n) - (A + B)| \leq |a_n - A| + |b_n - B| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \]
Therefore, for every \(\epsilon > 0\), there exists an \(N\) such that for all \(n \geq N\), we have:
\[ |(a_n + b_n) - (A + B)| < \epsilon \]
This proves that:
\[ \lim_{n \to \infty} (a_n + b_n) = A + B \]
Limit of a Product
To prove this theorem, we must show that for every \(\epsilon > 0\), there exists a natural number \(N\) such that:
\[ |(a_n \cdot b_n) - (A \cdot B)| < \epsilon \quad \text{for all} \ n \geq N. \]
Since \(\lim_{n \to \infty} a_n = A\), for every \(\epsilon > 0\), there exists a natural number \(N_1\) such that:
\[ |a_n - A| < \frac{\epsilon}{2 \cdot (|B| + 1)} \quad \text{for all} \ n \geq N_1. \]
Since \(\lim_{n \to \infty} b_n = B\), for every \(\epsilon > 0\), there exists a natural number \(N_2\) such that:
\[ |b_n - B| < \frac{\epsilon}{2 \cdot (|A| + 1)} \quad \text{for all} \ n \geq N_2. \]
We choose \(N\) as the maximum between \(N_1\) and \(N_2\):
\[ N = \max(N_1, N_2). \]
For all \(n \geq N\), we have both \(n \geq N_1\) and \(n \geq N_2\). Therefore:
\[ |a_n - A| < \frac{\epsilon}{2 \cdot (|B| + 1)} \quad \text{and} \quad |b_n - B| < \frac{\epsilon}{2 \cdot (|A| + 1)}. \]
Let us consider the difference:
\[ |(a_n \cdot b_n) - (A \cdot B)|. \]
We can rewrite this difference as:
\[ |(a_n \cdot b_n) - (A \cdot B)| = |a_n \cdot b_n - A \cdot B|. \]
We add and subtract \(A \cdot b_n\):
\[ |a_n \cdot b_n - A \cdot B| = |(a_n \cdot b_n - A \cdot b_n) + (A \cdot b_n - A \cdot B)|. \]
We use the triangle inequality:
\[ |(a_n \cdot b_n) - (A \cdot B)| \leq |a_n \cdot b_n - A \cdot b_n| + |A \cdot b_n - A \cdot B|. \]
We decompose the terms:
\[ |a_n \cdot b_n - A \cdot b_n| = |(a_n - A) \cdot b_n|. \]
Since:
\[ |a_n - A| < \frac{\epsilon}{2 \cdot (|B| + 1)} \]
moreover:
\[ |b_n| \leq |B| + 1 \text{ for } n \geq N, \]
we have:
\[ |(a_n - A) \cdot b_n| \leq |a_n - A| \cdot |b_n| < \frac{\epsilon}{2 \cdot (|B| + 1)} \cdot (|B| + 1) = \frac{\epsilon}{2}. \]
Similarly:
\[ |A \cdot b_n - A \cdot B| = |A \cdot (b_n - B)|. \]
Since:
\[ |b_n - B| < \frac{\epsilon}{2 \cdot (|A| + 1)}, \]
moreover:
\[ |A| \leq |A| + 1 \text{ for } n \geq N, \]
we have:
\[ |A \cdot (b_n - B)| \leq |A| \cdot |b_n - B| < (|A| + 1) \cdot \frac{\epsilon}{2 \cdot (|A| + 1)} = \frac{\epsilon}{2}. \]
We add the inequalities:
\[ |(a_n \cdot b_n) - (A \cdot B)| \leq |(a_n - A) \cdot b_n| + |A \cdot (b_n - B)| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \]
Therefore, for every \(\epsilon > 0\), there exists an \(N\) such that for all \(n \geq N\), we have:
\[ |(a_n \cdot b_n) - (A \cdot B)| < \epsilon. \]
This proves that:
\[ \lim_{n \to \infty} (a_n \cdot b_n) = A \cdot B. \]
Limit of a Quotient
Suppose we have two sequences \(a_n\) and \(b_n\) such that:
\[ \lim_{n \to \infty} a_n = A \quad \text{and} \quad \lim_{n \to \infty} b_n = B, \]
with \(B \neq 0\). We want to prove that:
\[ \lim_{n \to \infty} \frac{a_n}{b_n} = \frac{A}{B}. \]
This means that, for every \(\epsilon > 0\), there exists a natural number \(N\) such that, for all \(n \geq N\), we have:
\[ \left| \frac{a_n}{b_n} - \frac{A}{B} \right| < \epsilon. \]
We rewrite the difference as:
\[ \left| \frac{a_n}{b_n} - \frac{A}{B} \right| = \left| \frac{a_n \cdot B - A \cdot b_n}{b_n \cdot B} \right|. \]
To estimate this expression, we can use the triangle inequality and rewrite the numerator:
\[ |a_n \cdot B - A \cdot b_n| = |(a_n - A) \cdot B + A \cdot (B - b_n)|. \]
Applying the triangle inequality, we obtain:
\[ |(a_n - A) \cdot B + A \cdot (B - b_n)| \leq |(a_n - A) \cdot B| + |A \cdot (B - b_n)|. \]
Now we must estimate separately the two terms \( |(a_n - A) \cdot B| \) and \( |A \cdot (B - b_n)| \). To do this, we can choose two quantities \( \delta_1 \) and \( \delta_2 \), such that:
\[ \delta_1 + \delta_2 = \epsilon. \]
We impose the following conditions:
\[ |a_n - A| < \delta_1 \quad \text{and} \quad |b_n - B| < \delta_2. \]
Using these inequalities, we obtain for the first term:
\[ |(a_n - A) \cdot B| \leq |a_n - A| \cdot |B| < \delta_1 \cdot |B|. \]
For the second term:
\[ |A \cdot (B - b_n)| \leq |A| \cdot |B - b_n| < |A| \cdot \delta_2. \]
We now add the two terms:
\[ |a_n \cdot B - A \cdot b_n| \leq \delta_1 \cdot |B| + \delta_2 \cdot |A|. \]
Finally, we divide by \( |b_n \cdot B| \). Since for \(n\) sufficiently large \( |b_n| \geq \frac{|B|}{2} \), we can write:
\[ \frac{|a_n \cdot B - A \cdot b_n|}{|b_n \cdot B|} \leq \frac{\delta_1 \cdot |B| + \delta_2 \cdot |A|}{\frac{|B|^2}{2}}. \]
Simplifying the expression, we obtain:
\[ \frac{2(\delta_1 \cdot |B| + \delta_2 \cdot |A|)}{|B|^2}. \]
To ensure that the entire expression is less than \( \epsilon \), we can choose \( \delta_1 \) and \( \delta_2 \) so that they satisfy a proportional relationship. For example, we can choose:
\[ \delta_1 = \frac{\epsilon}{2} \quad \text{and} \quad \delta_2 = \frac{\epsilon}{2}. \]
In this way, we ensure that:
\[ \left| \frac{a_n}{b_n} - \frac{A}{B} \right| < \epsilon, \]
thus proving that:
\[ \lim_{n \to \infty} \frac{a_n}{b_n} = \frac{A}{B}. \]