real analysis question prove If limn→∞
an = a and limn→∞
bn = b, then limn→∞
(anbn) = ab.

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

---

### **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 \( n \geq N \),  
\[
|a_n b_n - ab| < \epsilon.
\]

---

### **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:

\[
|a_n b_n - ab| \leq |b_n| |a_n - a| + |a| |b_n - b|.
\]

---

### **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 \( n \geq N_2 \), \( |b_n - b| < \epsilon_2 \).

---

### **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:

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

---

### **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:
\[
\lim_{n \to \infty} (a_n b_n) = ab.
\]

---

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.

Prove: If lim(n→∞) an = a and lim(n→∞) bn = b, then lim(n→∞) (anbn) = ab.

Explore a proof in real analysis showing that if sequences converge to limits a and b, then their product converges to ab. This solution uses the epsilon-delta definition of limits and the triangle inequality for real sequences.

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