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.