To find the limit as $x$ approaches 4 for the expression

$\lim_{{x \to 4}} \frac{x^2 - 16}{x^2 - x - 12}$

we start by examining the form of the function at $x = 4$.

Step 1: Checking for an Indeterminate Form

When $x = 4$:

• The numerator $x^2 - 16 = 4^2 - 16 = 16 - 16 = 0$.
• The denominator $x^2 - x - 12 = 4^2 - 4 - 12 = 16 - 4 - 12 = 0$.

Thus, the limit is of the indeterminate form $\frac{0}{0}$, allowing us to apply L'Hôpital's Rule, which involves differentiating the numerator and the denominator separately.

Step 2: Applying L'Hôpital's Rule

1. Differentiate the numerator: $\frac{d}{dx}(x^2 - 16) = 2x$.
2. Differentiate the denominator: $\frac{d}{dx}(x^2 - x - 12) = 2x - 1$.

Now, we rewrite the limit as:

$\lim_{{x \to 4}} \frac{2x}{2x - 1}$

Step 3: Substitute $x = 4$

Substitute $x = 4$ into the new expression:

$\frac{2(4)}{2(4) - 1} = \frac{8}{8 - 1} = \frac{8}{7}$

Final Answer

Therefore,

$\lim_{{x \to 4}} \frac{x^2 - 16}{x^2 - x - 12} = \frac{8}{7}$

Would you like further details on any part of the process or have additional questions?

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1. Why is L'Hôpital's Rule applicable for the $\frac{0}{0}$ indeterminate form?
2. How do we verify that a limit exists after applying L'Hôpital’s Rule?
3. What are other types of indeterminate forms where L'Hôpital’s Rule applies?
4. How can we confirm the correctness of the differentiation steps?
5. Can limits involving L'Hôpital’s Rule always be simplified by factoring?

Tip: Always check if simple factoring can resolve the expression before applying L'Hôpital’s Rule; sometimes factoring avoids differentiation altogether.