1. **Problem:** Evaluate $$\lim_{x \to -1} 2e^{3x+4}$$ 2. **Formula and rules:** The exponential function $$e^x$$ is continuous everywhere, so $$\lim_{x \to a} e^{f(x)} = e^{\lim_{x \to a} f(x)}$$. 3. **Work:** Substitute $$x = -1$$ into the exponent: $$3(-1) + 4 = -3 + 4 = 1$$ 4. **Evaluate:** $$\lim_{x \to -1} 2e^{3x+4} = 2e^1 = 2e$$ --- 1. **Problem:** Evaluate $$\lim_{x \to 2} \log_3(4x + 1)$$ 2. **Formula and rules:** The logarithm function $$\log_a(x)$$ is continuous for $$x > 0$$, so we can substitute directly if the argument is positive. 3. **Work:** Substitute $$x = 2$$: $$4(2) + 1 = 8 + 1 = 9$$ 4. **Evaluate:** $$\lim_{x \to 2} \log_3(4x + 1) = \log_3(9)$$ Since $$9 = 3^2$$, then $$\log_3(9) = 2$$ --- 1. **Problem:** Evaluate $$\lim_{x \to \frac{\pi}{2}} \sec\left(x + \frac{\pi}{3}\right)$$ 2. **Formula and rules:** $$\sec(\theta) = \frac{1}{\cos(\theta)}$$. The limit exists if $$\cos\left(x + \frac{\pi}{3}\right) \neq 0$$ at the limit point. 3. **Work:** Substitute $$x = \frac{\pi}{2}$$: $$x + \frac{\pi}{3} = \frac{\pi}{2} + \frac{\pi}{3} = \frac{3\pi}{6} + \frac{2\pi}{6} = \frac{5\pi}{6}$$ 4. **Evaluate:** $$\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}$$ Therefore, $$\lim_{x \to \frac{\pi}{2}} \sec\left(x + \frac{\pi}{3}\right) = \frac{1}{-\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}}$$ --- 1. **Problem:** Evaluate $$\lim_{x \to -6} 3 \left(\frac{1}{2}\right)^{x+3}$$ 2. **Formula and rules:** Exponential functions $$a^x$$ with $$a > 0$$ are continuous everywhere. 3. **Work:** Substitute $$x = -6$$: $$x + 3 = -6 + 3 = -3$$ 4. **Evaluate:** $$3 \left(\frac{1}{2}\right)^{-3} = 3 \times 2^3 = 3 \times 8 = 24$$ --- 1. **Problem:** From the graph, find $$\lim_{x \to 2\pi} h(x)$$. 2. **Observation:** The graph value at $$x = 2\pi$$ is approximately 1. 3. **Evaluate:** $$\lim_{x \to 2\pi} h(x) = 1$$ --- 1. **Problem:** From the graph, find $$\lim_{x \to \frac{\pi}{2}} h(x)$$. 2. **Observation:** The graph value at $$x = \frac{\pi}{2}$$ is 0. 3. **Evaluate:** $$\lim_{x \to \frac{\pi}{2}} h(x) = 0$$