The operations on limits are of fundamental importance because they allow us to calculate the limit of a sum, a product, or a quotient by directly deducing it from the limits of individual sequences. These rules greatly simplify calculations, allowing us to analyze the behavior of complex functions without resorting to more elaborate methods.
Table of Contents
Limit of the 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 every \(n \geq N\) the inequalities hold:
\[ |a_n - A| < \frac{\epsilon}{2} \quad \text{and} \quad |b_n - B| < \frac{\epsilon}{2} \]
Now, for every \(n \geq N\):
\[ |(a_n + b_n) - (A + B)| \leq |a_n - A| + |b_n - B| \]
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 every \(n \geq N\), we have:
\[ |(a_n + b_n) - (A + B)| < \epsilon \]
Which proves that:
\[ \lim_{n \to \infty} (a_n + b_n) = A + B \]
Limit of the 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 every \(n \geq N\), we have both \(n \geq N_1\) and \(n \geq N_2\). Therefore:
\[ |a_n - A| < \frac{\epsilon}{2 \cdot (|B| + 1)} \quad \text{and} \quad |b_n - B| < \frac{\epsilon}{2 \cdot (|A| + 1)}. \]
Consider the difference:
\[ |(a_n \cdot b_n) - (A \cdot B)|. \]
We can rewrite this difference as:
\[ |(a_n \cdot b_n) - (A \cdot B)| = |a_n \cdot b_n - A \cdot B|. \]
Add and subtract \(A \cdot b_n\):
\[ |a_n \cdot b_n - A \cdot B| = |(a_n \cdot b_n - A \cdot b_n) + (A \cdot b_n - A \cdot B)|. \]
Using 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:
\[ |A \cdot (b_n - B)| \leq |A| \cdot |b_n - B| < (|A| + 1) \cdot \frac{\epsilon}{2 \cdot (|A| + 1)} = \frac{\epsilon}{2}. \]
Adding 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 every \(n \geq N\), we have:
\[ |(a_n \cdot b_n) - (A \cdot B)| < \epsilon. \]
Which proves that:
\[ \lim_{n \to \infty} (a_n \cdot b_n) = A \cdot B. \]
Limit of the 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 show 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 every \(n \geq N\), we have:
\[ \left| \frac{a_n}{b_n} - \frac{A}{B} \right| < \epsilon. \]
Let's 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 get:
\[ |(a_n - A) \cdot B + A \cdot (B - b_n)| \leq |(a_n - A) \cdot B| + |A \cdot (B - b_n)|. \]
Now we need to 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 get for the first term:
\[ |(a_n - A) \cdot B| \leq |a_n - A| \cdot |B| < \delta_1 \cdot |B|. \]
For the second term:
\[ |A \cdot (B - b_n)| \leq |A| \cdot |B - b_n| < |A| \cdot \delta_2. \]
Now let's sum 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 sufficiently large \(n\), \( |b_n| \geq \frac{|B|}{2} \), 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 get:
\[ \frac{2(\delta_1 \cdot |B| + \delta_2 \cdot |A|)}{|B|^2}. \]
To ensure that the entire expression is less than \( \epsilon \), we can choose \( \delta_1 \) and \( \delta_2 \) in such a way 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:
\[ \lim_{n \to \infty} \frac{a_n}{b_n} = \frac{A}{B}. \]