1) Problem: Find $$\lim_{x \to -\infty} \frac{3x}{x+2}$$ Step 1. For limits at infinity of rational functions, divide numerator and denominator by the highest power of $$x$$ in the denominator. Step 2. Divide numerator and denominator by $$x$$: $$\lim_{x \to -\infty} \frac{3x/x}{(x+2)/x} = \lim_{x \to -\infty} \frac{3}{1 + 2/x}$$ Step 3. As $$x \to -\infty$$, $$2/x \to 0$$, so the limit is $$\frac{3}{1+0} = 3$$. Answer: A) 3 2) Problem: Find $$\lim_{x \to -\infty} \frac{-3x}{x-1}$$ Step 1. Divide numerator and denominator by $$x$$: $$\lim_{x \to -\infty} \frac{-3x/x}{(x-1)/x} = \lim_{x \to -\infty} \frac{-3}{1 - 1/x}$$ Step 2. As $$x \to -\infty$$, $$1/x \to 0$$, so the limit is $$\frac{-3}{1-0} = -3$$. Answer: A) -3 3) Problem: Find $$\lim_{x \to -\infty} \frac{-x}{x-1}$$ Step 1. Divide numerator and denominator by $$x$$: $$\lim_{x \to -\infty} \frac{-x/x}{(x-1)/x} = \lim_{x \to -\infty} \frac{-1}{1 - 1/x}$$ Step 2. As $$x \to -\infty$$, $$1/x \to 0$$, so the limit is $$\frac{-1}{1-0} = -1$$. Answer: B) -1 4) Problem: Find $$\lim_{x \to -\pi} -\sec(x)$$ Step 1. Recall $$\sec(x) = \frac{1}{\cos(x)}$$. Step 2. Evaluate $$\cos(-\pi) = \cos(\pi) = -1$$ (cosine is even). Step 3. So $$-\sec(-\pi) = -\frac{1}{\cos(-\pi)} = -\frac{1}{-1} = 1$$. Answer: B) 1 5) Problem: Find $$\lim_{x \to -\frac{2\pi}{3}} \tan(x)$$ Step 1. Evaluate $$\tan\left(-\frac{2\pi}{3}\right)$$. Step 2. Use $$\tan(-\theta) = -\tan(\theta)$$, so $$\tan\left(-\frac{2\pi}{3}\right) = -\tan\left(\frac{2\pi}{3}\right)$$. Step 3. $$\tan\left(\frac{2\pi}{3}\right) = \tan\left(\pi - \frac{\pi}{3}\right) = -\tan\left(\frac{\pi}{3}\right) = -\sqrt{3}$$. Step 4. Therefore, $$\tan\left(-\frac{2\pi}{3}\right) = -(-\sqrt{3}) = \sqrt{3}$$. Answer: A) \sqrt{3}