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The given expression is:

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

### Step-by-step Solution:

1. **Substitute \(x = 2\) directly:**
   \[
   \frac{\tan((2)^2 - 4)}{\sin(2 - 2)} = \frac{\tan(4 - 4)}{\sin(0)} = \frac{\tan(0)}{\sin(0)} = \frac{0}{0}
   \]
   This results in the indeterminate form \(\frac{0}{0}\), so we can apply **L'Hôpital's Rule**.

2. **Differentiate the numerator and denominator:**
   - The derivative of the numerator \(\tan(x^2 - 4)\) using the chain rule is:
     \[
     \frac{d}{dx} \tan(x^2 - 4) = \sec^2(x^2 - 4) \cdot \frac{d}{dx}(x^2 - 4) = \sec^2(x^2 - 4) \cdot 2x
     \]
   - The derivative of the denominator \(\sin(x - 2)\) is:
     \[
     \frac{d}{dx} \sin(x - 2) = \cos(x - 2)
     \]

3. **Rewrite the limit with the derivatives:**
   \[
   \lim_{x \to 2} \frac{2x \sec^2(x^2 - 4)}{\cos(x - 2)}
   \]

4. **Substitute \(x = 2\) into the new expression:**
   - For the numerator:
     \[
     2(2) \sec^2((2)^2 - 4) = 4 \sec^2(0) = 4 \cdot 1 = 4
     \]
   - For the denominator:
     \[
     \cos(2 - 2) = \cos(0) = 1
     \]

5. **Final Result:**
   \[
   \lim_{x \to 2} \frac{4}{1} = 4
   \]

Thus, the limit is:

\[
\boxed{4}
\]

Would you like more details on any step?

Here are 5 related questions for practice:
1. How do you apply L'Hôpital's Rule when you encounter a different indeterminate form?
2. What is the derivative of \(\tan(u)\) and why does the chain rule apply here?
3. Can you find the limit of \(\lim_{x \to 0} \frac{\sin(x)}{x}\)?
4. What is the difference between indeterminate forms and undefined values?
5. How does L'Hôpital's Rule handle higher-order derivatives when the first derivatives still lead to \( \frac{0}{0} \)?

**Tip:** Always check if the limit can be computed directly before using L'Hôpital's Rule.

lim (x → 2) (tan(x² - 4) / sin(x - 2))

Solve the limit problem lim (x → 2) (tan(x² - 4) / sin(x - 2)) using L'Hôpital's Rule. This step-by-step solution covers derivatives, indeterminate forms, and trigonometric functions. Suitable for students in Grades 11-12 or early university.

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