https://r2.erweima.ai/mathbot/image/17270649954463slh6y2z.jpg

Let's work through the limits in the image one by one:

### 13. \(\lim_{x \to \infty} \frac{\sqrt{2x^2 - 1}}{3x}\)

1. Simplify the expression inside the square root:
   \[
   \sqrt{2x^2 - 1} = \sqrt{x^2(2 - \frac{1}{x^2})} = x\sqrt{2 - \frac{1}{x^2}}
   \]
   As \(x \to \infty\), \(\frac{1}{x^2} \to 0\), so \(\sqrt{2 - \frac{1}{x^2}} \to \sqrt{2}\).
   
   Therefore:
   \[
   \frac{\sqrt{2x^2 - 1}}{3x} \approx \frac{x\sqrt{2}}{3x} = \frac{\sqrt{2}}{3}
   \]
   
   So, the limit is:
   \[
   \lim_{x \to \infty} \frac{\sqrt{2x^2 - 1}}{3x} = \frac{\sqrt{2}}{3}
   \]

---

### 14. \(\lim_{x \to \infty} \frac{x - 5}{\sqrt{x^2 + 2}}\)

1. Factor out \(x\) in the denominator:
   \[
   \sqrt{x^2 + 2} = \sqrt{x^2(1 + \frac{2}{x^2})} = x\sqrt{1 + \frac{2}{x^2}}
   \]
   As \(x \to \infty\), \(\frac{2}{x^2} \to 0\), so \(\sqrt{1 + \frac{2}{x^2}} \to 1\).
   
2. Now the expression simplifies to:
   \[
   \frac{x - 5}{x\sqrt{1 + \frac{2}{x^2}}} \approx \frac{x}{x} = 1
   \]
   
   Therefore, the limit is:
   \[
   \lim_{x \to \infty} \frac{x - 5}{\sqrt{x^2 + 2}} = 1
   \]

---

### 15. \(\lim_{x \to \infty} \frac{3x + 1}{\sqrt{x^2 - 1}}\)

1. Simplify the denominator:
   \[
   \sqrt{x^2 - 1} = \sqrt{x^2(1 - \frac{1}{x^2})} = x\sqrt{1 - \frac{1}{x^2}}
   \]
   As \(x \to \infty\), \(\frac{1}{x^2} \to 0\), so \(\sqrt{1 - \frac{1}{x^2}} \to 1\).

2. Simplify the expression:
   \[
   \frac{3x + 1}{x\sqrt{1 - \frac{1}{x^2}}} \approx \frac{3x + 1}{x} = 3 + \frac{1}{x}
   \]
   As \(x \to \infty\), \(\frac{1}{x} \to 0\), so the limit is:
   \[
   \lim_{x \to \infty} \frac{3x + 1}{\sqrt{x^2 - 1}} = 3
   \]

---

### 16. \(\lim_{x \to \infty} \frac{5x^3 - 1}{\sqrt{2 + x^6}}\)

1. Simplify the denominator:
   \[
   \sqrt{2 + x^6} = x^3\sqrt{\frac{2}{x^6} + 1} = x^3\sqrt{1 + \frac{2}{x^6}}
   \]
   As \(x \to \infty\), \(\frac{2}{x^6} \to 0\), so \(\sqrt{1 + \frac{2}{x^6}} \to 1\).

2. Simplify the expression:
   \[
   \frac{5x^3 - 1}{x^3\sqrt{1 + \frac{2}{x^6}}} \approx \frac{5x^3}{x^3} = 5
   \]
   
   Therefore, the limit is:
   \[
   \lim_{x \to \infty} \frac{5x^3 - 1}{\sqrt{2 + x^6}} = 5
   \]

---

### Summary of Results:
- \( \lim_{x \to \infty} \frac{\sqrt{2x^2 - 1}}{3x} = \frac{\sqrt{2}}{3} \)
- \( \lim_{x \to \infty} \frac{x - 5}{\sqrt{x^2 + 2}} = 1 \)
- \( \lim_{x \to \infty} \frac{3x + 1}{\sqrt{x^2 - 1}} = 3 \)
- \( \lim_{x \to \infty} \frac{5x^3 - 1}{\sqrt{2 + x^6}} = 5 \)

Would you like more details on any of these solutions? 

Here are 5 follow-up questions for practice:

1. How would the limit change in problem 13 if the coefficient of \(x^2\) under the square root was different?
2. What happens to the limit in problem 14 if the numerator was \(x^2 - 5\) instead of \(x - 5\)?
3. Can you think of a scenario where the denominator in problem 15 dominates the numerator? How would that change the result?
4. How would you approach the limit in problem 16 if the denominator had a different exponent for \(x\)?
5. Can you generalize the method of simplifying limits involving square roots?

**Tip**: When working with limits involving square roots at infinity, it helps to factor out the highest power of \(x\) to simplify the expression.

Calculate the limits for various algebraic expressions as x approaches infinity.

Explore detailed solutions to problems involving limits of ratios of polynomials and radicals as x approaches infinity. This guide includes a step-by-step breakdown using calculus properties and simplification techniques, suitable for first-year calculus students.

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