Let's go over these limit problems one by one.

39. $\lim_{x \to 0} \frac{\sin 3x}{x}$

Using the known limit property $\lim_{x \to 0} \frac{\sin x}{x} = 1$, we can rewrite: $\lim_{x \to 0} \frac{\sin 3x}{x} = 3 \lim_{x \to 0} \frac{\sin 3x}{3x} = 3 \cdot 1 = 3$ Answer: 3

40. $\lim_{x \to 0} \frac{\sin 4x}{\sin 6x}$

We can apply the small angle approximation for limits involving sine: $\lim_{x \to 0} \frac{\sin 4x}{\sin 6x} = \lim_{x \to 0} \frac{4x}{6x} = \frac{4}{6} = \frac{2}{3}$ Answer: $\frac{2}{3}$

41. $\lim_{\theta \to 0} \frac{\tan 6\theta}{\sin 2\theta}$

Using the approximation $\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$ and $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$, we have: $\lim_{\theta \to 0} \frac{\tan 6\theta}{\sin 2\theta} = \frac{6}{2} = 3$ Answer: 3

42. $\lim_{\theta \to 0} \frac{\cos \theta - 1}{\sin \theta}$

This is a standard limit, which can be found using L'Hôpital's Rule (as it's of the form $\frac{0}{0}$): $\lim_{\theta \to 0} \frac{\cos \theta - 1}{\sin \theta} = \lim_{\theta \to 0} \frac{-\sin \theta}{\cos \theta} = 0$ Answer: 0

43. $\lim_{\theta \to 0} \frac{\sin(\cos \theta)}{\sec \theta}$

For small $\theta$, $\cos \theta \approx 1$, so: $\lim_{\theta \to 0} \frac{\sin(\cos \theta)}{\sec \theta} = \frac{\sin 1}{1} = \sin 1$ Answer: $\sin 1$

44. $\lim_{t \to 0} \frac{\sin^2 3t}{t^2}$

This can be rewritten as: $\lim_{t \to 0} \frac{\sin^2 3t}{t^2} = \lim_{t \to 0} \left( \frac{\sin 3t}{t} \right)^2 = 9$ Answer: 9

45. $\lim_{\theta \to 0} \frac{\sin \theta}{\theta + \tan \theta}$

For small $\theta$, both $\sin \theta \approx \theta$ and $\tan \theta \approx \theta$: $\lim_{\theta \to 0} \frac{\sin \theta}{\theta + \tan \theta} = \lim_{\theta \to 0} \frac{\theta}{2\theta} = \frac{1}{2}$ Answer: $\frac{1}{2}$

46. $\lim_{x \to 0} \frac{\sin(x^2)}{x}$

Since $\sin(x^2) \approx x^2$ for small $x$: $\lim_{x \to 0} \frac{\sin(x^2)}{x} = \lim_{x \to 0} \frac{x^2}{x} = \lim_{x \to 0} x = 0$ Answer: 0

47. $\lim_{x \to \pi/4} \frac{1 - \tan x}{\sin x - \cos x}$

At $x = \frac{\pi}{4}$, both $\tan \frac{\pi}{4} = 1$, $\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$, so: $\lim_{x \to \pi/4} \frac{1 - 1}{\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}} = \frac{0}{0}$ This suggests applying L'Hôpital's Rule. After differentiation: $\lim_{x \to \pi/4} \frac{-\sec^2 x}{\cos x + \sin x} = \frac{-2}{\sqrt{2}} = -\sqrt{2}$ Answer: $-\sqrt{2}$

48. $\lim_{x \to 1} \frac{\sin(x - 1)}{x^2 + x - 2}$

Factor the denominator: $x^2 + x - 2 = (x - 1)(x + 2)$, so: $\lim_{x \to 1} \frac{\sin(x - 1)}{(x - 1)(x + 2)} = \lim_{x \to 1} \frac{1}{x + 2} = \frac{1}{3}$ Answer: $\frac{1}{3}$

Would you like further clarification or more details on any of these? Here are 5 related questions to consider:

1. How do you apply L'Hôpital's Rule in more complex limits?
2. What are the common approximations used for trigonometric functions near zero?
3. How does the chain rule affect limit computations involving compositions of functions?
4. What are the best strategies to factor complex polynomials in limit problems?
5. How does Taylor expansion help in solving trigonometric limits?

Tip: Always check if the limit is an indeterminate form (like $\frac{0}{0}$) before using advanced techniques like L'Hôpital's Rule.