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Let's break down and solve the limit problems in the image you uploaded. The problems are as follows:

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**8. Compute the following limits:**

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### (a) \(\lim\limits_{x \to 2} \frac{x^2 - 2x}{x^3 - 8}\)

This limit involves factoring both the numerator and the denominator. First, notice that the denominator can be factored as a difference of cubes:

- \(x^3 - 8 = (x - 2)(x^2 + 2x + 4)\)

The numerator can be factored as:

- \(x^2 - 2x = x(x - 2)\)

Now, we rewrite the expression:

\[
\lim_{x \to 2} \frac{x(x - 2)}{(x - 2)(x^2 + 2x + 4)}
\]

We can cancel out the \((x - 2)\) terms:

\[
\lim_{x \to 2} \frac{x}{x^2 + 2x + 4}
\]

Substitute \(x = 2\) into the remaining expression:

\[
\frac{2}{2^2 + 2(2) + 4} = \frac{2}{4 + 4 + 4} = \frac{2}{12} = \frac{1}{6}
\]

Thus, the limit is:

\[
\boxed{\frac{1}{6}}
\]

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### (b) \(\lim\limits_{h \to 0} \frac{\sqrt[3]{27 + h} - 3}{h}\)

The hint suggests that we make the substitution \( u = \sqrt[3]{27 + h} \). Let's analyze the expression.

We are dealing with the limit of a difference quotient, which is characteristic of a derivative. Specifically, this is the derivative of the function \( f(x) = \sqrt[3]{x} \) at \(x = 27\). 

We know that the derivative of \( f(x) = x^{1/3} \) is:

\[
f'(x) = \frac{1}{3}x^{-2/3}
\]

At \(x = 27\), the derivative is:

\[
f'(27) = \frac{1}{3}(27)^{-2/3} = \frac{1}{3 \cdot 3^2} = \frac{1}{27}
\]

Therefore, the limit is:

\[
\boxed{\frac{1}{27}}
\]

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### (c) \(\lim\limits_{x \to 1} \frac{x^n - 1}{x - 1}\), where \(n\) is a natural number.

This limit represents the difference quotient of the function \( f(x) = x^n \) at \(x = 1\). This is a standard limit, and it can be evaluated using L'Hopital's Rule or by recognizing it as the derivative of \( f(x) = x^n \) at \(x = 1\).

The derivative of \( f(x) = x^n \) is:

\[
f'(x) = nx^{n-1}
\]

At \(x = 1\), the derivative is:

\[
f'(1) = n \cdot 1^{n-1} = n
\]

Therefore, the limit is:

\[
\boxed{n}
\]

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Do you need any further explanation or details on these problems?

### Follow-up Questions:
1. Can you explain the process behind L'Hopital's Rule and when it's used?
2. How is the derivative of a cube root function calculated in general?
3. Why do we factor in part (a) and how does it help in solving the limit?
4. How can substitution simplify limit problems, as seen in part (b)?
5. Can you elaborate on other cases where the difference quotient represents a derivative?

### Tip:
When approaching limit problems, always check if direct substitution works first. If not, consider factoring, simplifying, or applying L'Hopital's Rule for indeterminate forms.

Compute the following limits: a) lim x → 2 (x² - 2x) / (x³ - 8), b) lim h → 0 (³√(27 + h) - 3) / h, c) lim x → 1 (x^n - 1) / (x - 1) where n is a natural number.

This page covers limit problems involving factoring and derivatives, using methods like L'Hopital's Rule and algebraic simplifications. Learn to solve problems such as lim x → 2 (x² - 2x) / (x³ - 8) and lim h → 0 (³√(27 + h) - 3) / h. Suitable for advanced high school or early college calculus students.

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