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The image contains the following expression, with the instruction to evaluate the limit as \( x \) approaches 2:

\[
\lim_{x \to 2} \frac{32 - x^5}{2x^6 - 128}
\]

Let's begin by simplifying the expression and evaluating the limit.

### Step 1: Factor the denominator
Notice that the denominator \( 2x^6 - 128 \) can be factored:

\[
2x^6 - 128 = 2(x^6 - 64) = 2(x^3 - 8)(x^3 + 8)
\]

Since \( 64 = 2^6 \), we recognize \( x^6 - 64 \) as a difference of cubes.

### Step 2: Evaluate the limit at \( x = 2 \)
We now substitute \( x = 2 \) directly into the expression:

- The numerator: \( 32 - x^5 = 32 - 2^5 = 32 - 32 = 0 \)
- The denominator: \( 2x^6 - 128 = 2(2^6) - 128 = 2(64) - 128 = 128 - 128 = 0 \)

Both the numerator and the denominator evaluate to zero, which gives a \( \frac{0}{0} \) indeterminate form. This suggests we need to apply **L'Hopital's Rule**.

### Step 3: Apply L'Hopital's Rule
We take the derivative of the numerator and denominator:

- Derivative of the numerator \( 32 - x^5 \) is \( -5x^4 \).
- Derivative of the denominator \( 2x^6 - 128 \) is \( 12x^5 \).

Now the limit becomes:

\[
\lim_{x \to 2} \frac{-5x^4}{12x^5}
\]

### Step 4: Simplify and evaluate the new limit
Simplify the expression:

\[
\frac{-5x^4}{12x^5} = \frac{-5}{12x}
\]

Now substitute \( x = 2 \):

\[
\frac{-5}{12(2)} = \frac{-5}{24}
\]

### Final Answer:
\[
\lim_{x \to 2} \frac{32 - x^5}{2x^6 - 128} = \frac{-5}{24}
\]

Would you like further explanation or have any questions?

Here are 5 related questions:
1. What is L'Hopital's Rule, and when do you apply it?
2. How do you factor a difference of cubes like \( x^6 - 64 \)?
3. What happens if a limit gives a different indeterminate form, such as \( \infty/\infty \)?
4. Can limits always be evaluated using substitution?
5. How do derivatives help in evaluating limits with indeterminate forms?

**Tip**: When evaluating limits, always check if direct substitution works before applying more advanced techniques like factoring or L'Hopital's Rule.

Evaluate the limit of the expression (32 - x^5) / (2x^6 - 128) as x approaches 2.

Evaluate the limit of (32 - x^5) / (2x^6 - 128) as x approaches 2 using L'Hopital's Rule. This detailed solution covers indeterminate forms, factoring, and differentiation, suitable for Grades 11-12 students.

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