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Let's analyze and solve each of these limits one by one.

### 1. \(\lim_{{x \to 1}} \frac{{x^4 - 1}}{{x^3 - 1}}\)

**Step-by-Step Solution:**
1. Notice that both the numerator and the denominator will be zero if we directly substitute \(x = 1\), resulting in an indeterminate form \( \frac{0}{0} \).
2. Factor both the numerator and denominator:
   - \(x^4 - 1 = (x^2 + 1)(x - 1)(x + 1)\)
   - \(x^3 - 1 = (x - 1)(x^2 + x + 1)\)
3. Cancel out the \((x - 1)\) terms from the numerator and the denominator:
   \[
   \frac{{x^4 - 1}}{{x^3 - 1}} = \frac{{(x^2 + 1)(x + 1)}}{{x^2 + x + 1}}
   \]
4. Now, substitute \(x = 1\):
   \[
   \frac{{(1^2 + 1)(1 + 1)}}{{1^2 + 1 + 1}} = \frac{{(1 + 1)(2)}}{{3}} = \frac{4}{3}
   \]
**Answer:** \(\frac{4}{3}\)

---

### 2. \(\lim_{{x \to 1}} \frac{{e^{x - 1} - 1}}{{8x - 8}}\)

**Step-by-Step Solution:**
1. Again, if we substitute \(x = 1\), we get an indeterminate form \( \frac{0}{0} \).
2. Simplify the denominator: \(8x - 8 = 8(x - 1)\).
3. Rewrite the limit:
   \[
   \lim_{{x \to 1}} \frac{{e^{x - 1} - 1}}{{8(x - 1)}}
   \]
4. Apply L'Hôpital's Rule by differentiating the numerator and the denominator:
   - The derivative of \(e^{x - 1} - 1\) is \(e^{x - 1}\).
   - The derivative of \(8(x - 1)\) is \(8\).
5. Substitute \(x = 1\):
   \[
   \frac{e^{1 - 1}}{8} = \frac{1}{8}
   \]
**Answer:** \(\frac{1}{8}\)

---

### 3. \(\lim_{{x \to 9}} \frac{{x^2 - 9^2}}{{\sin(\pi x)}}\)

**Step-by-Step Solution:**
1. Notice that both the numerator and the denominator are zero when \(x = 9\), which is an indeterminate form.
2. Factor the numerator as a difference of squares:
   \[
   x^2 - 9^2 = (x - 9)(x + 9)
   \]
   So the expression becomes:
   \[
   \frac{{(x - 9)(x + 9)}}{{\sin(\pi x)}}
   \]
3. Substitute \(x = 9\) directly, but observe that \(\sin(\pi x) = \sin(9\pi) = 0\). So we need to consider the behavior of this limit as \(x\) approaches 9. We can use the fact that as \(x \to 9\), \(\sin(\pi x) \approx \pi(x - 9)\).
4. Simplify the expression using this approximation:
   \[
   \frac{{(x - 9)(x + 9)}}{{\pi(x - 9)}} = \frac{{x + 9}}{\pi}
   \]
5. Substitute \(x = 9\):
   \[
   \frac{{9 + 9}}{\pi} = \frac{18}{\pi}
   \]
**Answer:** \(\frac{18}{\pi}\)

---

### Summary of Answers
1. \(\lim_{{x \to 1}} \frac{{x^4 - 1}}{{x^3 - 1}} = \frac{4}{3}\)
2. \(\lim_{{x \to 1}} \frac{{e^{x - 1} - 1}}{{8x - 8}} = \frac{1}{8}\)
3. \(\lim_{{x \to 9}} \frac{{x^2 - 9^2}}{{\sin(\pi x)}} = \frac{18}{\pi}\)

Would you like a detailed explanation of any specific step?

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#### Related Questions:
1. How do you apply L'Hôpital's Rule in other indeterminate forms?
2. How do approximations help in finding trigonometric limits?
3. What are other techniques for solving limits without direct substitution?
4. How can you factor more complex polynomials to simplify limits?
5. How does the Taylor series expansion apply to limits involving exponential functions?

#### Tip:
When you encounter indeterminate forms like \( \frac{0}{0} \), L'Hôpital's Rule or factoring can often simplify the problem effectively.

Find the following limits:
1. \(\lim_{{x \to 1}} \frac{{x^4 - 1}}{{x^3 - 1}}\)
2. \(\lim_{{x \to 1}} \frac{{e^{x - 1} - 1}}{{8x - 8}}\)
3. \(\lim_{{x \to 9}} \frac{{x^2 - 9^2}}{{\sin(\pi x)}}\)

Solve limits involving indeterminate forms through techniques such as L'Hôpital's Rule, factoring, and trigonometric approximations. Detailed solutions include step-by-step explanations for problems like \(\lim_{{x \to 1}} \frac{{x^4 - 1}}{{x^3 - 1}}\), \(\lim_{{x \to 1}} \frac{{e^{x - 1} - 1}}{{8x - 8}}\), and \(\lim_{{x \to 9}} \frac{{x^2 - 9^2}}{{\sin(\pi x)}}\).

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