1. **Problem Q.4 (i):** Use the Sandwich Theorem to find $$\lim_{x \to 0} f(x)$$ given $$\sqrt{5 - 2x^2} \leq f(x) \leq \sqrt{5 - x^2}$$. 2. The Sandwich Theorem states if $$g(x) \leq f(x) \leq h(x)$$ and $$\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$$, then $$\lim_{x \to a} f(x) = L$$. 3. Calculate the limits of the bounding functions: $$\lim_{x \to 0} \sqrt{5 - 2x^2} = \sqrt{5 - 0} = \sqrt{5}$$ $$\lim_{x \to 0} \sqrt{5 - x^2} = \sqrt{5 - 0} = \sqrt{5}$$ 4. Since both limits are equal, by the Sandwich Theorem: $$\lim_{x \to 0} f(x) = \sqrt{5}$$. 5. **Problem Q.4 (ii):** Find $$\lim_{x \to 0} g(x)$$ given $$2 - x^2 \leq g(x) \leq 2 \cos x$$. 6. Calculate the limits of bounding functions: $$\lim_{x \to 0} (2 - x^2) = 2 - 0 = 2$$ $$\lim_{x \to 0} 2 \cos x = 2 \cdot 1 = 2$$ 7. By the Sandwich Theorem: $$\lim_{x \to 0} g(x) = 2$$. 8. **Problem Q.5 (i):** Find $$\lim_{x \to 1^+} \sqrt{\frac{x-1}{x+2}}$$. 9. As $$x \to 1^+$$, numerator $$x-1 \to 0^+$$ and denominator $$x+2 \to 3$$. 10. So inside the root, $$\frac{x-1}{x+2} \to \frac{0^+}{3} = 0^+$$. 11. Therefore, limit is $$\sqrt{0^+} = 0$$. 12. **Problem Q.5 (ii):** Find $$\lim_{x \to -2^+} \frac{(x+3)|x+2|}{x+2}$$. 13. For $$x \to -2^+$$, $$x+2 > 0$$, so $$|x+2| = x+2$$. 14. Substitute: $$\frac{(x+3)(x+2)}{x+2} = x+3$$ (since $$x+2 \neq 0$$ near limit). 15. Evaluate limit: $$\lim_{x \to -2^+} (x+3) = -2 + 3 = 1$$. 16. **Problem Q.6 (i):** Find $$\lim_{x \to 0} \frac{1}{3x}$$. 17. As $$x \to 0$$, denominator $$3x \to 0$$. 18. From the right $$x \to 0^+$$, $$\frac{1}{3x} \to +\infty$$; from the left $$x \to 0^-$$, $$\frac{1}{3x} \to -\infty$$. 19. So the limit does not exist (infinite limit with different signs). 20. **Problem Q.6 (ii):** Find $$\lim_{x \to 0^-} \frac{5}{2x}$$. 21. As $$x \to 0^-$$, denominator $$2x \to 0^-$$. 22. So $$\frac{5}{2x} \to -\infty$$. 23. **Problem Q.6 (iii):** Find $$\lim_{x \to -5^-} \frac{3x}{2x + 10}$$. 24. Substitute $$x = -5$$: Denominator: $$2(-5) + 10 = -10 + 10 = 0$$. 25. For $$x \to -5^-$$, $$2x + 10 \to 0^-$$ (since $$x < -5$$). 26. Numerator $$3x \to 3(-5) = -15$$. 27. So limit is $$\frac{-15}{0^-} = +\infty$$. 28. **Problem Q.6 (iv):** Find $$\lim_{\theta \to 0} (1 + \csc \theta)$$. 29. Recall $$\csc \theta = \frac{1}{\sin \theta}$$. 30. As $$\theta \to 0$$, $$\sin \theta \to 0$$, so $$\csc \theta \to \pm \infty$$ depending on direction. 31. Since $$\sin \theta \to 0^+$$ from right, $$\csc \theta \to +\infty$$; from left, $$\csc \theta \to -\infty$$. 32. Therefore, limit does not exist. 33. **Problem Q.1 (i):** Find $$\lim_{x \to 1} \frac{-1}{3x - 1}$$. 34. Substitute $$x=1$$: $$3(1) - 1 = 2$$. 35. So limit is $$\frac{-1}{2} = -\frac{1}{2}$$. 36. **Problem Q.1 (ii):** Find $$\lim_{x \to \frac{1}{3}} (3x - 1)$$. 37. Substitute $$x=\frac{1}{3}$$: $$3 \cdot \frac{1}{3} - 1 = 1 - 1 = 0$$. 38. **Problem Q.1 (iii):** Find $$\lim_{x \to \frac{\pi}{2}} x \sin x$$. 39. Substitute $$x=\frac{\pi}{2}$$: $$\frac{\pi}{2} \cdot \sin \frac{\pi}{2} = \frac{\pi}{2} \cdot 1 = \frac{\pi}{2}$$. 40. **Problem Q.1 (iv):** Find $$\lim_{x \to \pi} \frac{\cos x}{1 - \pi}$$. 41. Substitute $$x=\pi$$: $$\cos \pi = -1$$. 42. Denominator is constant $$1 - \pi$$. 43. So limit is $$\frac{-1}{1 - \pi}$$. 44. **Problem Q.2 (i)(a):** Average rate of change of $$g(x) = x^2$$ over $$[-1,1]$$. 45. Formula: $$\frac{g(b) - g(a)}{b - a}$$. 46. Calculate: $$\frac{1^2 - (-1)^2}{1 - (-1)} = \frac{1 - 1}{2} = 0$$. 47. **Problem Q.2 (i)(b):** Average rate of change over $$[-2,0]$$. 48. Calculate: $$\frac{0^2 - (-2)^2}{0 - (-2)} = \frac{0 - 4}{2} = -2$$. 49. **Problem Q.2 (ii)(a):** Average rate of change of $$g(t) = 2 + \cos t$$ over $$[0, \pi]$$. 50. Calculate: $$\frac{(2 + \cos \pi) - (2 + \cos 0)}{\pi - 0} = \frac{(2 - 1) - (2 + 1)}{\pi} = \frac{1 - 3}{\pi} = \frac{-2}{\pi}$$. 51. **Problem Q.2 (ii)(b):** Average rate of change over $$[-\pi, \pi]$$. 52. Calculate: $$\frac{(2 + \cos \pi) - (2 + \cos (-\pi))}{\pi - (-\pi)} = \frac{(2 - 1) - (2 - 1)}{2\pi} = \frac{1 - 1}{2\pi} = 0$$. 53. **Problem Q.3 (i):** Find $$\lim_{x \to -1} 3(2x - 1)^2$$. 54. Substitute $$x = -1$$: $$3(2(-1) - 1)^2 = 3(-2 - 1)^2 = 3(-3)^2 = 3 \cdot 9 = 27$$. 55. **Problem Q.3 (ii):** Find $$\lim_{h \to 0} \frac{3}{\sqrt{3h + 1} + 1}$$. 56. Substitute $$h=0$$: $$\frac{3}{\sqrt{1} + 1} = \frac{3}{1 + 1} = \frac{3}{2}$$. 57. **Problem Q.3 (iii):** Find $$\lim_{x \to 0} \frac{x^2 + x - 2}{x^2 - 1}$$. 58. Substitute $$x=0$$: Numerator: $$0 + 0 - 2 = -2$$ Denominator: $$0 - 1 = -1$$ 59. Limit is $$\frac{-2}{-1} = 2$$. 60. **Problem Q.3 (iv):** Find $$\lim_{x \to 4} \frac{4x - x^2}{2 - \sqrt{x}}$$. 61. Substitute $$x=4$$: Numerator: $$4(4) - 4^2 = 16 - 16 = 0$$ Denominator: $$2 - 2 = 0$$ 62. Use algebraic manipulation: $$\frac{4x - x^2}{2 - \sqrt{x}} = \frac{x(4 - x)}{2 - \sqrt{x}}$$. 63. Multiply numerator and denominator by conjugate $$2 + \sqrt{x}$$: $$\frac{x(4 - x)(2 + \sqrt{x})}{(2 - \sqrt{x})(2 + \sqrt{x})} = \frac{x(4 - x)(2 + \sqrt{x})}{4 - x}$$. 64. Cancel $$4 - x$$: $$x(2 + \sqrt{x})$$. 65. Substitute $$x=4$$: $$4(2 + 2) = 4 \cdot 4 = 16$$. 66. **Problem Q.3 (v):** Find $$\lim_{x \to -1} \frac{\sqrt{x^2 + 8} - 3}{x + 1}$$. 67. Substitute $$x = -1$$: Numerator: $$\sqrt{1 + 8} - 3 = 3 - 3 = 0$$ Denominator: $$-1 + 1 = 0$$ 68. Use conjugate multiplication: $$\frac{\sqrt{x^2 + 8} - 3}{x + 1} \cdot \frac{\sqrt{x^2 + 8} + 3}{\sqrt{x^2 + 8} + 3} = \frac{x^2 + 8 - 9}{(x + 1)(\sqrt{x^2 + 8} + 3)} = \frac{x^2 - 1}{(x + 1)(\sqrt{x^2 + 8} + 3)}$$. 69. Factor numerator: $$\frac{(x - 1)(x + 1)}{(x + 1)(\sqrt{x^2 + 8} + 3)} = \frac{x - 1}{\sqrt{x^2 + 8} + 3}$$. 70. Substitute $$x = -1$$: $$\frac{-1 - 1}{\sqrt{1 + 8} + 3} = \frac{-2}{3 + 3} = \frac{-2}{6} = -\frac{1}{3}$$. **Final answers:** Q.4 (i): $$\sqrt{5}$$ Q.4 (ii): $$2$$ Q.5 (i): $$0$$ Q.5 (ii): $$1$$ Q.6 (i): Limit does not exist (infinite with different signs) Q.6 (ii): $$-\infty$$ Q.6 (iii): $$+\infty$$ Q.6 (iv): Limit does not exist Q.1 (i): $$-\frac{1}{2}$$ Q.1 (ii): $$0$$ Q.1 (iii): $$\frac{\pi}{2}$$ Q.1 (iv): $$\frac{-1}{1 - \pi}$$ Q.2 (i)(a): $$0$$ Q.2 (i)(b): $$-2$$ Q.2 (ii)(a): $$-\frac{2}{\pi}$$ Q.2 (ii)(b): $$0$$ Q.3 (i): $$27$$ Q.3 (ii): $$\frac{3}{2}$$ Q.3 (iii): $$2$$ Q.3 (iv): $$16$$ Q.3 (v): $$-\frac{1}{3}$$