Let's work through the requested limit problems from number 8 to 14 step by step.

Problem 8:

$\lim_{\theta \to 0} \frac{\sin 2\theta}{\sin \theta}$ Using the double-angle identity for sine, $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$, we rewrite the limit: $\lim_{\theta \to 0} \frac{2\sin(\theta)\cos(\theta)}{\sin(\theta)}$ Simplifying: $\lim_{\theta \to 0} 2\cos(\theta)$ Since $\cos(0) = 1$, the limit is: $\boxed{2}$

---

Problem 9:

$\lim_{\theta \to 0} \frac{\cot(\pi \theta) \sin(\theta)}{2\sec(\theta)}$ We know that $\cot(x) = \frac{\cos(x)}{\sin(x)}$ and $\sec(x) = \frac{1}{\cos(x)}$, so substitute these into the limit: $\lim_{\theta \to 0} \frac{\frac{\cos(\pi \theta)}{\sin(\pi \theta)} \sin(\theta)}{2\frac{1}{\cos(\theta)}}$ Simplifying: $\lim_{\theta \to 0} \frac{\cos(\pi \theta)\sin(\theta)\cos(\theta)}{2\sin(\pi \theta)}$ As $\theta \to 0$, both $\sin(\theta)$ and $\sin(\pi \theta)$ tend to 0, but using L'Hopital's rule or small-angle approximations for sine and cosine, we can find: $\boxed{0}$

---

Problem 10:

$\lim_{t \to 0} \frac{\sin^2(3t)}{2t}$ First, use the identity $\sin^2(x) = (1 - \cos(2x))/2$ and substitute, or alternatively, factor the expression as follows: $\lim_{t \to 0} \frac{\sin(3t)}{t} \cdot \frac{\sin(3t)}{2}$ Using $\lim_{t \to 0} \frac{\sin(3t)}{t} = 3$, the limit becomes: $3 \cdot 0 = \boxed{0}$

---

Problem 11:

$\lim_{t \to 0} \frac{\tan^2(3t)}{2t}$ We can use the approximation $\tan(x) \approx x$ when $x$ is small. Thus: $\lim_{t \to 0} \frac{(3t)^2}{2t} = \lim_{t \to 0} \frac{9t^2}{2t} = \lim_{t \to 0} \frac{9t}{2} = 0$ Thus, the limit is: $\boxed{0}$

---

Problem 12:

$\lim_{\theta \to 0} \frac{\tan(2\theta)}{\sin(2\theta) - 1}$ Using $\tan(x) \approx x$ and $\sin(x) \approx x$ for small angles: $\lim_{\theta \to 0} \frac{2\theta}{2\theta - 1}$ At $\theta = 0$, the denominator becomes $-1$, so the limit is: $\boxed{0}$

---

Problem 13:

$\lim_{t \to 0} \frac{\sin(3t) + 4t}{t \sec(t)}$ We use the small-angle approximation for sine and secant ($\sec(t) \approx 1 + t^2/2$): $\lim_{t \to 0} \frac{3t + 4t}{t(1)} = \lim_{t \to 0} \frac{7t}{t} = 7$ Thus, the limit is: $\boxed{7}$

---

Problem 14:

$\lim_{\theta \to 0} \frac{\sin^2(\theta)}{\theta^2}$ We recognize this as a standard limit form. Since $\lim_{\theta \to 0} \frac{\sin(\theta)}{\theta} = 1$, we have: $\lim_{\theta \to 0} \frac{\sin^2(\theta)}{\theta^2} = \left( \lim_{\theta \to 0} \frac{\sin(\theta)}{\theta} \right)^2 = 1^2 = 1$ Thus, the limit is: $\boxed{1}$

---

Would you like further details or have any questions about the steps?

Here are 5 related questions to explore:

1. What is L'Hopital's Rule and how can it be applied to limits?
2. How do approximations for small angles ($\sin(x) \approx x$, $\tan(x) \approx x$) help simplify limit problems?
3. What is the significance of the standard limit $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$?
4. How do trigonometric identities assist in solving complex limit problems?
5. When do we need to use Taylor series expansions in evaluating limits?

Tip: When evaluating limits involving trigonometric functions, always check if small-angle approximations can simplify the expression!