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Let's solve each limit one by one.

### Problem Analysis

**Instructions:**
- Use \( \text{inf} \) for \( +\infty \)
- Use \( -\text{inf} \) for \( -\infty \)
- Use \( \text{dne} \) if the limit does not exist

### Solutions

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**(a) \( \lim_{x \to \infty} \frac{8x^7 + 2x + 175}{13x^7 + 1} \)**

1. **Dominant term**: Since \( x \to \infty \), we consider the highest power term in both numerator and denominator.
   - Numerator: \( 8x^7 \)
   - Denominator: \( 13x^7 \)
2. Simplify the expression by dividing both the numerator and denominator by \( x^7 \):
   \[
   \lim_{x \to \infty} \frac{8x^7 + 2x + 175}{13x^7 + 1} = \lim_{x \to \infty} \frac{8 + \frac{2}{x^6} + \frac{175}{x^7}}{13 + \frac{1}{x^7}} = \frac{8}{13}
   \]
3. **Answer**: \( \frac{8}{13} \).

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**(b) \( \lim_{x \to \infty} \frac{e^x}{800x^5 + \sin(x)} \)**

1. **Dominant term**: As \( x \to \infty \), \( e^x \) grows much faster than any polynomial term \( x^5 \) or the bounded term \( \sin(x) \).
2. Thus, the numerator \( e^x \) will tend to infinity faster than the denominator.
3. **Answer**: \( \text{inf} \).

---

**(c) \( \lim_{x \to \infty} \frac{\sqrt{x} \log(x^{70})}{x^2} \)**

1. **Rewrite**:
   \[
   \frac{\sqrt{x} \log(x^{70})}{x^2} = \frac{\sqrt{x} \cdot 70 \log(x)}{x^2} = \frac{70 \cdot x^{1/2} \log(x)}{x^2}
   \]
2. **Simplify**: Rewrite \( x^{1/2} \) in terms of \( x^2 \):
   \[
   = \frac{70 \log(x)}{x^{3/2}}
   \]
3. As \( x \to \infty \), \( x^{3/2} \) grows faster than \( \log(x) \), so the limit approaches \( 0 \).
4. **Answer**: \( 0 \).

---

**(d) \( \lim_{x \to \infty} \left( \sqrt{x^2 + 1} - \sqrt{x^2 - 1} \right) \)**

1. **Multiply by the conjugate**:
   \[
   \left( \sqrt{x^2 + 1} - \sqrt{x^2 - 1} \right) \cdot \frac{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}}{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}}
   \]
   gives
   \[
   = \frac{(x^2 + 1) - (x^2 - 1)}{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}} = \frac{2}{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}}
   \]
2. **Simplify**: As \( x \to \infty \),
   \[
   = \frac{2}{x\sqrt{1 + \frac{1}{x^2}} + x\sqrt{1 - \frac{1}{x^2}}} = \frac{2}{2x} = \frac{1}{x}
   \]
3. As \( x \to \infty \), \( \frac{1}{x} \to 0 \).
4. **Answer**: \( 0 \).

---

### Final Answers:

1. (a) \( \frac{8}{13} \)
2. (b) \( \text{inf} \)
3. (c) \( 0 \)
4. (d) \( 0 \)

Would you like any of these steps explained further?

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### Relative Questions

1. How do we determine the dominant terms in polynomial expressions when calculating limits at infinity?
2. Why is \( e^x \) dominant over polynomial expressions as \( x \to \infty \)?
3. When is it beneficial to use the conjugate method in limit problems?
4. How does logarithmic growth compare to polynomial and exponential growth as \( x \to \infty \)?
5. What does it mean for a limit to "not exist," and how does that differ from approaching \( \infty \) or \( -\infty \)?

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### Tip

When evaluating limits as \( x \to \infty \), focus on the highest power term in each part of the expression, as lower power terms become negligible.

Compute the following limits:
(a) \( \lim_{{x \to \infty}} \frac{{8x^7 + 2x + 175}}{{13x^7 + 1}} \)
(b) \( \lim_{{x \to \infty}} \frac{{e^x}}{{800x^5 + \sin(x)}} \)
(c) \( \lim_{{x \to \infty}} \frac{{\sqrt{x} \log(x^{70})}}{{x^2}} \)
(d) \( \lim_{{x \to \infty}} \left( \sqrt{x^2 + 1} - \sqrt{x^2 - 1} \right) \)

Learn how to compute limits at infinity using dominant term analysis and simplification techniques. This detailed solution covers different approaches for evaluating limits, including the behavior of exponential, polynomial, and square root functions. Suitable for undergraduate calculus students.

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