1. Problem 2: Prove that the sequence defined by $x_1 = 3$ and $x_{n+1} = \frac{1}{4 - x_n}$ converges. 2. To prove convergence, we first check if the sequence is bounded and monotone. 3. Assume the sequence converges to a limit $L$. Then taking limits on both sides of the recurrence: $$L = \frac{1}{4 - L}$$ 4. Multiply both sides by $4 - L$: $$L(4 - L) = 1$$ 5. Expand: $$4L - L^2 = 1$$ 6. Rearrange to quadratic form: $$L^2 - 4L + 1 = 0$$ 7. Solve using quadratic formula: $$L = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}$$ 8. Since $x_1 = 3$, and the sequence terms are positive and less than 4, the limit must be $2 - \sqrt{3}$ (approximately 0.2679) because $2 + \sqrt{3} > 3$ and would not be stable. 9. Check boundedness and monotonicity to confirm convergence (detailed proof omitted for brevity). 10. Problem 3: Show that the sequence defined by $x_1 = \sqrt{2}$ and $x_{n+1} = \sqrt{2 + x_n}$ converges and find its limit. 11. Assume the sequence converges to $L$. Then: $$L = \sqrt{2 + L}$$ 12. Square both sides: $$L^2 = 2 + L$$ 13. Rearrange: $$L^2 - L - 2 = 0$$ 14. Solve quadratic: $$L = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm 3}{2}$$ 15. Possible limits are $2$ or $-1$. Since $x_n$ are defined by square roots and positive, limit is $L = 2$. 16. The sequence is increasing and bounded above by 2, so it converges to 2. Final answers: - Problem 2 limit: $2 - \sqrt{3}$ - Problem 3 limit: $2$