1. Problem: Find $\lim_{x \to +\infty} \cos \left(\frac{1}{x}\right)$. As $x \to +\infty$, $\frac{1}{x} \to 0$. Since cosine is continuous, $$\lim_{x \to +\infty} \cos \left(\frac{1}{x}\right) = \cos(0) = 1.$$ 2. Problem: Find $\lim_{x \to +\infty} \sin \left(\frac{\pi x}{2 - 3x}\right)$. Rewrite the argument: $$\frac{\pi x}{2 - 3x} = \frac{\pi x}{-3x + 2} = \frac{\pi x}{-3x(1 - \frac{2}{3x})} = \frac{\pi}{-3} \cdot \frac{1}{1 - \frac{2}{3x}}.$$ As $x \to +\infty$, $\frac{2}{3x} \to 0$, so $$\frac{\pi x}{2 - 3x} \to \frac{\pi}{-3} = -\frac{\pi}{3}.$$ Therefore, $$\lim_{x \to +\infty} \sin \left(\frac{\pi x}{2 - 3x}\right) = \sin \left(-\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}.$$ 3. Problem: Find $\lim_{\theta \to 0} \frac{\sin 3\theta}{\theta}$. Rewrite as $$\frac{\sin 3\theta}{\theta} = 3 \cdot \frac{\sin 3\theta}{3\theta}.$$ As $\theta \to 0$, $\frac{\sin 3\theta}{3\theta} \to 1$, so $$\lim_{\theta \to 0} \frac{\sin 3\theta}{\theta} = 3 \cdot 1 = 3.$$ 4. Problem: Find $\lim_{h \to 0} \frac{\sin h}{2h}$. Rewrite as $$\frac{\sin h}{2h} = \frac{1}{2} \cdot \frac{\sin h}{h}.$$ As $h \to 0$, $\frac{\sin h}{h} \to 1$, so $$\lim_{h \to 0} \frac{\sin h}{2h} = \frac{1}{2} \cdot 1 = \frac{1}{2}.$$ 5. Problem: Find $\lim_{x \to 0} \frac{x^2 - 3 \sin x}{x}$. Rewrite numerator: $$\frac{x^2 - 3 \sin x}{x} = \frac{x^2}{x} - 3 \cdot \frac{\sin x}{x} = x - 3 \cdot \frac{\sin x}{x}.$$ As $x \to 0$, $x \to 0$ and $\frac{\sin x}{x} \to 1$, so $$\lim_{x \to 0} \frac{x^2 - 3 \sin x}{x} = 0 - 3 \cdot 1 = -3.$$ 6. Problem: Find $\lim_{x \to 0} \frac{2 - \cos 3x - \cos 4x}{x}$. Use expansions: $$\cos kx \approx 1 - \frac{(kx)^2}{2}$$ near 0. So numerator $$2 - \cos 3x - \cos 4x \approx 2 - \left(1 - \frac{9x^2}{2}\right) - \left(1 - \frac{16x^2}{2}\right) = 2 - 1 + \frac{9x^2}{2} - 1 + \frac{16x^2}{2} = \frac{25x^2}{2}.$$ Divide by $x$: $$\frac{2 - \cos 3x - \cos 4x}{x} \approx \frac{25x^2/2}{x} = \frac{25x}{2}.$$ As $x \to 0$, this goes to 0. 7. Problem: Find $\lim_{\theta \to 0^+} \frac{\sin \theta}{\theta^2}$. As $\theta \to 0$, $\sin \theta \approx \theta$, so $$\frac{\sin \theta}{\theta^2} \approx \frac{\theta}{\theta^2} = \frac{1}{\theta}.$$ As $\theta \to 0^+$, $\frac{1}{\theta} \to +\infty$. 8. Problem: Find $\lim_{\theta \to 0} \frac{\sin^2 \theta}{\theta}$. Rewrite as $$\frac{\sin^2 \theta}{\theta} = \sin \theta \cdot \frac{\sin \theta}{\theta}.$$ As $\theta \to 0$, $\sin \theta \to 0$ and $\frac{\sin \theta}{\theta} \to 1$, so $$\lim_{\theta \to 0} \frac{\sin^2 \theta}{\theta} = 0 \cdot 1 = 0.$$