The given limit problem is:

$\lim_{x \to -2} \frac{(x^2 - 4) \cdot \tan(x+2)}{\sin^2(x+2)}$

Step 1: Factorize $x^2 - 4$

The expression $x^2 - 4$ is a difference of squares, so it can be factorized as:

$x^2 - 4 = (x - 2)(x + 2)$

Thus, the limit becomes:

$\lim_{x \to -2} \frac{(x - 2)(x + 2) \cdot \tan(x+2)}{\sin^2(x+2)}$

Step 2: Substitution

Let $u = x + 2$, which implies that as $x \to -2$, $u \to 0$. Substituting $u$ into the limit:

$\lim_{u \to 0} \frac{(u - 4) \cdot \tan(u)}{\sin^2(u)}$

Step 3: Analyze Limit Behavior

• As $u \to 0$, $\tan(u) \approx u$ and $\sin(u) \approx u$ (from small angle approximations).
• Using these approximations:

$\lim_{u \to 0} \frac{u \cdot u}{u^2} = \lim_{u \to 0} 1 = 1$

Therefore, the limit of the entire expression is:

$\boxed{4}$

Would you like further explanation or details?

Here are five related questions to expand your understanding:

1. How do you handle limits involving trigonometric functions?
2. What is the small angle approximation for trigonometric functions like sine and tangent?
3. How does L'Hôpital's Rule apply to indeterminate forms in limits?
4. What is the general approach to solving limits involving factorization?
5. Can you solve limits without using trigonometric approximations?

Tip: For limits involving trigonometric functions near 0, small angle approximations such as $\sin(x) \approx x$ and $\tan(x) \approx x$ are very useful.