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Let's solve the given limit problems step by step.

### Exercise 11:

**1.** \(\lim_{x \to +\infty} \sqrt{3x^2 + 2}\)
- As \(x \to +\infty\), the dominant term in \(3x^2 + 2\) is \(3x^2\).
- \[
\sqrt{3x^2 + 2} \approx \sqrt{3x^2} = \sqrt{3}x
\]
Thus, 
\[
\lim_{x \to +\infty} \sqrt{3x^2 + 2} = +\infty.
\]

**2.** \(\lim_{x \to +\infty} \frac{x^3 - 5x + 2}{x^4}\)
- Divide numerator and denominator by \(x^4\):
  \[
  \lim_{x \to +\infty} \frac{\frac{x^3}{x^4} - \frac{5x}{x^4} + \frac{2}{x^4}}{1} = \lim_{x \to +\infty} \left(\frac{1}{x} - \frac{5}{x^3} + \frac{2}{x^4}\right) = 0.
  \]

**3.** \(\lim_{x \to +\infty} \frac{7x^3 + 4x}{x^4}\)
- Divide numerator and denominator by \(x^4\):
  \[
  \lim_{x \to +\infty} \frac{\frac{7x^3}{x^4} + \frac{4x}{x^4}}{1} = \lim_{x \to +\infty} \left(\frac{7}{x} + \frac{4}{x^3}\right) = 0.
  \]

### Exercise 12:

**1.** \(\lim_{x \to +\infty} \sqrt{5x - 4} - 1\)
- For large \(x\), the dominant term in \(\sqrt{5x - 4}\) is \(\sqrt{5x}\). So:
  \[
  \lim_{x \to +\infty} \left(\sqrt{5x - 4} - 1\right) \approx \lim_{x \to +\infty} \left(\sqrt{5x} - 1\right) = +\infty.
  \]

**2.** \(\lim_{x \to +\infty} \frac{3x + 4}{\sqrt{3x + 4}}\)
- As \(x \to +\infty\), both the numerator and denominator grow large. Approximate both terms:
  \[
  \lim_{x \to +\infty} \frac{3x + 4}{\sqrt{3x + 4}} \approx \lim_{x \to +\infty} \frac{3x}{\sqrt{3x}} = \lim_{x \to +\infty} \sqrt{3x} = +\infty.
  \]

### Exercise 13:

**1.** \(\lim_{x \to 1^+} \frac{x + 3}{x - 1}\)
- As \(x \to 1^+\), \(x - 1\) approaches 0 from the positive side, making the denominator very small. Thus, the limit goes to \(+\infty\).

**2.** \(\lim_{x \to 2} \frac{2x + 5}{(x - 2)^2}\)
- As \(x \to 2\), the numerator \(2x + 5 = 9\), and the denominator \((x - 2)^2 \to 0\). Since the denominator is squared, the limit is:
  \[
  \lim_{x \to 2} \frac{9}{(x - 2)^2} = +\infty.
  \]

**3.** \(\lim_{x \to +\infty} \frac{3x^2 + 1}{\sqrt{x^4 + 1}}\)
- For large \(x\), the dominant terms are \(3x^2\) in the numerator and \(x^4\) in the denominator:
  \[
  \lim_{x \to +\infty} \frac{3x^2 + 1}{\sqrt{x^4 + 1}} \approx \lim_{x \to +\infty} \frac{3x^2}{x^2} = 3.
  \]

Let me know if you need more details on any of these limits.

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### Related Questions:
1. How do we solve limits involving radicals like \(\sqrt{x^2 - 1}\)?
2. What is L'Hopital's Rule, and when should it be applied in limit problems?
3. How do you handle indeterminate forms in limits like \(\frac{0}{0}\)?
4. How do you determine limits as \(x\) approaches negative infinity?
5. Can limits at infinity result in finite values? How?

### Tip:
When solving limits at infinity, always check which terms dominate as \(x\) becomes large.

This lesson demonstrates solving limits involving polynomials and radicals as x approaches infinity. Concepts covered include dominant terms, polynomial division, and approximations in radical expressions. Suitable for students in Grades 10-12.

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