1. **Problem 1a:** Find the limit $$\lim_{x \to 3} \sqrt{x+1} - \sqrt{3x+7}$$ 2. To solve this limit without L'Hopital's rule, we use the conjugate to simplify: $$\lim_{x \to 3} \left(\sqrt{x+1} - \sqrt{3x+7}\right) = \lim_{x \to 3} \frac{(\sqrt{x+1} - \sqrt{3x+7})(\sqrt{x+1} + \sqrt{3x+7})}{\sqrt{x+1} + \sqrt{3x+7}}$$ 3. Simplify the numerator using difference of squares: $$= \lim_{x \to 3} \frac{(x+1) - (3x+7)}{\sqrt{x+1} + \sqrt{3x+7}} = \lim_{x \to 3} \frac{x + 1 - 3x - 7}{\sqrt{x+1} + \sqrt{3x+7}} = \lim_{x \to 3} \frac{-2x - 6}{\sqrt{x+1} + \sqrt{3x+7}}$$ 4. Substitute $x=3$ directly: $$= \frac{-2(3) - 6}{\sqrt{3+1} + \sqrt{3(3)+7}} = \frac{-6 - 6}{\sqrt{4} + \sqrt{16}} = \frac{-12}{2 + 4} = \frac{-12}{6} = -2$$ --- 5. **Problem 1b:** Find the limit $$\lim_{x \to 0} \frac{x^2 + \sin 2x}{x}$$ 6. Split the limit: $$= \lim_{x \to 0} \frac{x^2}{x} + \lim_{x \to 0} \frac{\sin 2x}{x} = \lim_{x \to 0} x + \lim_{x \to 0} \frac{\sin 2x}{x}$$ 7. The first limit is straightforward: $$\lim_{x \to 0} x = 0$$ 8. For the second limit, use the standard limit $\lim_{x \to 0} \frac{\sin ax}{x} = a$: $$\lim_{x \to 0} \frac{\sin 2x}{x} = 2$$ 9. Therefore, the total limit is: $$0 + 2 = 2$$ **Final answers:** - a) $$-2$$ - b) $$2$$