1. **State the problems:** Find the limits: $$\lim_{\theta \to 0} \frac{\tan \theta}{\theta^2 \cot 36}$$ and $$\lim_{\theta \to 0} \frac{\theta \cot 4\theta}{\sin^2 \theta \cot^2 2\theta}$$ 2. **Recall important formulas and rules:** - As $\theta \to 0$, $\tan \theta \approx \theta$. - As $\theta \to 0$, $\sin \theta \approx \theta$. - $\cot x = \frac{1}{\tan x}$. - Constants like $\cot 36$ are just numbers. 3. **Solve the first limit:** $$\lim_{\theta \to 0} \frac{\tan \theta}{\theta^2 \cot 36} = \lim_{\theta \to 0} \frac{\tan \theta}{\theta^2} \cdot \frac{1}{\cot 36}$$ Using $\tan \theta \approx \theta$ near zero: $$= \lim_{\theta \to 0} \frac{\theta}{\theta^2} \cdot \frac{1}{\cot 36} = \lim_{\theta \to 0} \frac{1}{\theta} \cdot \frac{1}{\cot 36}$$ As $\theta \to 0$, $\frac{1}{\theta} \to \infty$, so the limit diverges to infinity. **Answer:** The first limit does not exist (diverges to infinity). 4. **Solve the second limit:** $$\lim_{\theta \to 0} \frac{\theta \cot 4\theta}{\sin^2 \theta \cot^2 2\theta}$$ Rewrite cotangents: $$= \lim_{\theta \to 0} \frac{\theta \cdot \frac{1}{\tan 4\theta}}{(\sin \theta)^2 \cdot \left(\frac{1}{\tan 2\theta}\right)^2} = \lim_{\theta \to 0} \frac{\theta}{\tan 4\theta} \cdot \frac{\tan^2 2\theta}{\sin^2 \theta}$$ Use approximations near zero: - $\tan 4\theta \approx 4\theta$ - $\tan 2\theta \approx 2\theta$ - $\sin \theta \approx \theta$ Substitute: $$= \lim_{\theta \to 0} \frac{\theta}{4\theta} \cdot \frac{(2\theta)^2}{\theta^2} = \lim_{\theta \to 0} \frac{1}{4} \cdot \frac{4\theta^2}{\theta^2} = \frac{1}{4} \cdot 4 = 1$$ **Answer:** The second limit equals 1. **Final answers:** - First limit: Does not exist (diverges to infinity). - Second limit: 1.