The document you uploaded contains three exercises on calculus and algebra. Here’s an overview of each exercise:

Exercise 1 (12 Points)

1. Limits:

   • (a) \lim_{x \to 3} \frac{x^2 + 2x + 10 - 4\sqrt{x^4 + 6x - 1}}
   • (b) $\lim_{x \to +\infty} \frac{x^2 + x + \sqrt{x^2 - 3} - 2x}{x}$
   • (c) $\lim_{x \to \infty} \frac{\sqrt{x} - 3}{\sqrt{x} - 3}$
   • (d) $\lim_{x \to 1} \frac{\sqrt[3]{3x + 24} - 3}{x - 1}$
   • (e) $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$
   • (f) $\lim_{x \to 1} \frac{\sqrt{x - 1} - \sqrt{x - 1}}{x}$

2. Simplification:

   • (a) $\frac{\sqrt[3]{27} \times \sqrt{9} \times \sqrt{3}}{\sqrt{81} \times \sqrt{9}}$
   • (b) $\frac{\sqrt[5]{8} \times \sqrt[4]{16} \times \sqrt[6]{64}}{\sqrt[3]{2} \times \sqrt[2]{4}}$

3. Solve Equations:

   • (a) Solve for $x$ in $\sqrt{1 - x + x + 1} = \sqrt{2}$.
   • (b) Solve for $x$ in $\sqrt{(x + 3)^2} + \sqrt{(3 - x)^2} = 2\sqrt{9 - x^2}$.

4. Continuity:

   • Prove that the function f(x) = \frac{\sqrt{x^2 + 2x + 3} + \frac{x - 1}{x^2 + 2x + 3}} is continuous over $\mathbb{R}$.

Exercise 2 (2 Points)

Given a piecewise function $f(x)$ defined as: