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 the 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:

\[ |a_n - A| < \epsilon \quad \text{for every} \ n \geq N_1. \]

Similarly, \(\lim_{n \to \infty} b_n = B\) means that for every \(\epsilon > 0\), there exists a natural number \(N_2\) such that:

\[ |b_n - B| < \epsilon \quad \text{for every} \ n \geq N_2. \]

Now, we need to show that:

\[ \lim_{n \to \infty} (a_n + b_n) = A + B. \]

To prove this, we need to show that for every \(\epsilon > 0\), there exists a natural number \(N\) such that:

\[ |(a_n + b_n) - (A + B)| < \epsilon \quad \text{for every} \ n \geq N. \]

Consider the inequality:

\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:

\[ |a_n - A| < \frac{\epsilon}{2} \quad \text{for every} \ n \geq N_1. \]

Similarly, since \(\lim_{n \to \infty} b_n = B\), there exists a natural number \(N_2\) such that:

\[ |b_n - B| < \frac{\epsilon}{2} \quad \text{for every} \ n \geq N_2. \]

If we choose \(N = \max(N_1, N_2)\), we have that for every \(n \geq N\), the inequalities:

\[ |a_n - A| < \frac{\epsilon}{2} \quad \text{and} \quad |b_n - B| < \frac{\epsilon}{2} \]

hold. 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 get:

\[ |(a_n + b_n) - (A + B)| \leq |a_n - A| + |b_n - B| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \]

Therefore, for every \(\epsilon > 0\), there exists an \(N\) such that for every \(n \geq N\), we have:

\[ |(a_n + b_n) - (A + B)| < \epsilon. \]

This proves that:

\[ \lim_{n \to \infty} (a_n + b_n) = A + B. \]

Limit of a Product

To prove this theorem, we need to show that for every \(\epsilon > 0\), there exists a natural number \(N\) such that:

\[ |(a_n \cdot b_n) - (A \cdot B)| < \epsilon \quad \text{for every} \ n \geq N. \]

Since \(\lim_{n \to \infty} a_n = A\), for every \(\epsilon > 0\), there exists a natural number \(N_1\) such that:

\[ |a_n - A| < \frac{\epsilon}{2 \cdot (|B| + 1)} \quad \text{for every} \ n \geq N_1. \]

Since \(\lim_{n \to \infty} b_n = B\), for every \(\epsilon > 0\), there exists a natural number \(N_2\) such that:

\[ |b_n - B| < \frac{\epsilon}{2 \cdot (|A| + 1)} \quad \text{for every} \ n \geq N_2. \]

Choose \(N = \max(N_1, N_2)\), so that for every \(n \geq N\), both \(n \geq N_1\) and \(n \geq N_2\) hold. Thus:

\[ |a_n - A| < \frac{\epsilon}{2 \cdot (|B| + 1)} \quad \text{and} \quad |b_n - B| < \frac{\epsilon}{2 \cdot (|A| + 1)}. \]

Now, 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:

\[ |(a_n \cdot b_n) - (A \cdot B)| \leq |a_n \cdot b_n - A \cdot b_n| + |A \cdot b_n - A \cdot B|. \]

Break down the terms:

\[ |a_n \cdot b_n - A \cdot b_n| = |(a_n - A) \cdot b_n|. \]

Since:

\[ |a_n - A| < \frac{\epsilon}{2 \cdot (|B| + 1)}, \]

and:

\[ |b_n| \leq |B| + 1 \text{ for } n \geq N, \]

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:

\[ |b_n - B| < \frac{\epsilon}{2 \cdot (|A| + 1)}, \]

and:

\[ |A| \leq |A| + 1, \]

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 two inequalities together:

\[ |(a_n \cdot b_n) - (A \cdot B)| \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \]

Therefore, for every \(\epsilon > 0\), there exists an \(N\) such that for every \(n \geq N\), we have:

\[ |(a_n \cdot b_n) - (A \cdot B)| < \epsilon. \]

This proves that:

\[ \lim_{n \to \infty} (a_n \cdot b_n) = A \cdot B. \]

Limit of the Ratio

Let’s 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 every \(n \geq N\), we have:

\[ \left| \frac{a_n}{b_n} - \frac{A}{B} \right| < \epsilon. \]

We can 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 the two terms \( |(a_n - A) \cdot B| \) and \( |A \cdot (B - b_n)| \). To do so, we 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, for the first term, we get:

\[ |(a_n - A) \cdot B| \leq |a_n - A| \cdot |B| < \delta_1 \cdot |B|. \]

For the second term:

\[ |A \cdot (B - b_n)| \leq |A| \cdot |B - b_n| < |A| \cdot \delta_2. \]

Summing the two terms:

\[ |a_n \cdot B - A \cdot b_n| \leq \delta_1 \cdot |B| + \delta_2 \cdot |A|. \]

Finally, 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:

\[ \frac{2(\delta_1 \cdot |B| + \delta_2 \cdot |A|)}{|B|^2}. \]

To ensure the entire expression is less than \( \epsilon \), we can choose \( \delta_1 \) and \( \delta_2 \) such 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:

\[ \left| \frac{a_n}{b_n} - \frac{A}{B} \right| < \epsilon, \]

proving that:

\[ \lim_{n \to \infty} \frac{a_n}{b_n} = \frac{A}{B}. \]