To prove:
If $\lim_{n \to \infty} a_n = a$ and $\lim_{n \to \infty} b_n = b$, then $\lim_{n \to \infty} (a_n b_n) = ab$.

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Proof:

We will use the ε-δ definition of limits for sequences.

Since $\lim_{n \to \infty} a_n = a$, for any $\epsilon > 0$, there exists an integer $N_1$ such that for all $n \geq N_1$,
$|a_n - a| < \epsilon_1.$

Similarly, since $\lim_{n \to \infty} b_n = b$, for any $\epsilon > 0$, there exists an integer $N_2$ such that for all $n \geq N_2$,
$|b_n - b| < \epsilon_2.$

Now, we need to show that $\lim_{n \to \infty} (a_n b_n) = ab$.
That is, for any $\epsilon > 0$, we need to find an integer $N$ such that for all $n \geq N$,
$|a_n b_n - ab| < \epsilon.$

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Step 1: Manipulate $|a_n b_n - ab|$

$|a_n b_n - ab| = |a_n b_n - a b_n + a b_n - ab|.$ Using the triangle inequality:

$|a_n b_n - ab| \leq |a_n b_n - a b_n| + |a b_n - ab|.$

This can be simplified as:

$|a_n b_n - ab| \leq |b_n| |a_n - a| + |a| |b_n - b|.$

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Step 2: Control the terms

Since $\lim_{n \to \infty} b_n = b$, the sequence $\{b_n\}$ is bounded. Let’s say there exists a constant $M > 0$ such that for all $n$,
$|b_n| \leq M.$ Now, given $\epsilon > 0$, choose $\epsilon_1$ and $\epsilon_2$ such that:

$\epsilon_1 = \frac{\epsilon}{2M}, \quad \epsilon_2 = \frac{\epsilon}{2|a| + 1}.$

By the definition of limits, there exist integers $N_1$ and $N_2$ such that:

• For all $n \geq N_1$, $|a_n - a| < \epsilon_1$.
• For all $n \geq N_2$, $|b_n - b| < \epsilon_2$.

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Step 3: Final Bound

For $n \geq \max(N_1, N_2)$, we have:

$|a_n b_n - ab| \leq M \cdot \epsilon_1 + |a| \cdot \epsilon_2.$

Substitute the values of $\epsilon_1$ and $\epsilon_2$:

$|a_n b_n - ab| \leq M \cdot \frac{\epsilon}{2M} + |a| \cdot \frac{\epsilon}{2|a| + 1}.$

This simplifies to:

$|a_n b_n - ab| \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$

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Conclusion

Since for any $\epsilon > 0$, we found an $N = \max(N_1, N_2)$ such that for all $n \geq N$,
$|a_n b_n - ab| < \epsilon,$ we conclude that: $\lim_{n \to \infty} (a_n b_n) = ab.$

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Do you want further details or have any questions?

Here are 5 related questions for further exploration:

1. How does the triangle inequality help in limit proofs?
2. Can you generalize this result to complex sequences?
3. How does the boundedness of a sequence affect limit proofs?
4. Is it possible to prove this result using the epsilon-delta definition in a different way?
5. How would the proof change if one of the sequences converged to 0?

Tip: In real analysis, breaking down expressions with triangle inequalities is a common strategy to handle differences involving multiple terms.