https://file.aiquickdraw.com/mathbot/file/17298573041996wy8fqxp.jpg

Let's go through the limit problems one by one:

### 13. \(\lim_{x \to 2} \frac{x^3 - 8}{x - 2}\)
Notice that \(x^3 - 8\) can be factored as \((x - 2)(x^2 + 2x + 4)\). So the expression becomes:
\[
\frac{(x - 2)(x^2 + 2x + 4)}{x - 2}
\]
For \(x \neq 2\), the terms \(x - 2\) cancel out, leaving:
\[
\lim_{x \to 2} (x^2 + 2x + 4)
\]
Substituting \(x = 2\), we get:
\[
2^2 + 2(2) + 4 = 4 + 4 + 4 = 12
\]
So, the answer is:
\[
\boxed{12}
\]

### 14. \(\lim_{x \to 3} \frac{x^4 - 81}{x - 3}\)
Here, \(x^4 - 81\) is a difference of squares, which factors as:
\[
x^4 - 81 = (x^2 - 9)(x^2 + 9) = (x - 3)(x + 3)(x^2 + 9)
\]
Thus, the limit becomes:
\[
\lim_{x \to 3} \frac{(x - 3)(x + 3)(x^2 + 9)}{x - 3}
\]
Canceling \(x - 3\) (for \(x \neq 3\)), we get:
\[
\lim_{x \to 3} (x + 3)(x^2 + 9)
\]
Substitute \(x = 3\):
\[
(3 + 3)(3^2 + 9) = 6(9 + 9) = 6 \times 18 = 108
\]
So, the answer is:
\[
\boxed{108}
\]

### 15. \(\lim_{x \to 1} \frac{x^9 - 1}{x - 1}\)
Again, use factoring. \(x^9 - 1\) is a difference of powers, and can be factored as:
\[
x^9 - 1 = (x - 1)(x^8 + x^7 + \cdots + x + 1)
\]
So the expression becomes:
\[
\frac{(x - 1)(x^8 + x^7 + \cdots + x + 1)}{x - 1}
\]
Canceling \(x - 1\) gives:
\[
\lim_{x \to 1} (x^8 + x^7 + \cdots + x + 1)
\]
Substitute \(x = 1\):
\[
1^8 + 1^7 + \cdots + 1 + 1 = 1 + 1 + \cdots + 1 = 9
\]
So, the answer is:
\[
\boxed{9}
\]

### 16. \(\lim_{x \to 1} \frac{x^4 - 1}{x - 1}\)
Factor \(x^4 - 1\) as:
\[
x^4 - 1 = (x^2 - 1)(x^2 + 1) = (x - 1)(x + 1)(x^2 + 1)
\]
Thus, the limit becomes:
\[
\frac{(x - 1)(x + 1)(x^2 + 1)}{x - 1}
\]
Canceling \(x - 1\), we have:
\[
\lim_{x \to 1} (x + 1)(x^2 + 1)
\]
Substitute \(x = 1\):
\[
(1 + 1)(1^2 + 1) = 2 \times 2 = 4
\]
So, the answer is:
\[
\boxed{4}
\]

### 17. \(\lim_{x \to 9} \frac{\sqrt{x} - 3}{x - 9}\)
This is an indeterminate form \(\frac{0}{0}\), so we rationalize the numerator by multiplying by the conjugate:
\[
\frac{\sqrt{x} - 3}{x - 9} \times \frac{\sqrt{x} + 3}{\sqrt{x} + 3} = \frac{x - 9}{(x - 9)(\sqrt{x} + 3)} = \frac{1}{\sqrt{x} + 3}
\]
Now, substitute \(x = 9\):
\[
\frac{1}{\sqrt{9} + 3} = \frac{1}{3 + 3} = \frac{1}{6}
\]
So, the answer is:
\[
\boxed{\frac{1}{6}}
\]

### 18. \(\lim_{x \to 64} \frac{x^{1/3} - 4}{x - 64}\)
This is also indeterminate \(\frac{0}{0}\), so we can use the limit definition of the derivative. The expression is essentially the derivative of \(f(x) = x^{1/3}\) at \(x = 64\). Using the power rule for derivatives, \(f'(x) = \frac{1}{3}x^{-2/3}\).

Thus, at \(x = 64\), the derivative is:
\[
f'(64) = \frac{1}{3} \times 64^{-2/3} = \frac{1}{3} \times \frac{1}{16} = \frac{1}{48}
\]
So, the answer is:
\[
\boxed{\frac{1}{48}}
\]

Would you like further details or explanations on any of these solutions?

Here are five follow-up questions:
1. How do you apply the limit definition of the derivative to solve limits involving roots?
2. What are the factoring techniques used for higher degree polynomials in limits?
3. Why do indeterminate forms like \(\frac{0}{0}\) require special techniques like factoring or rationalizing?
4. How does L'Hôpital's Rule apply to limits and when should it be used?
5. Can you explain the concept of limits in real-world applications?

**Tip**: Always check for opportunities to factor expressions when solving limits algebraically. It often simplifies the process significantly!

Learn how to solve limits involving factoring, indeterminate forms, and rationalization. This detailed step-by-step guide covers the limits of polynomials and roots, suitable for high school students studying calculus.

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