1. The problem is to find the limit: $$\lim_{x \to -2} \frac{4 - x^2}{2 + x}$$. 2. We start by checking if direct substitution is possible by plugging in $x = -2$: $$\frac{4 - (-2)^2}{2 + (-2)} = \frac{4 - 4}{0} = \frac{0}{0}$$ which is an indeterminate form. 3. Since direct substitution gives $\frac{0}{0}$, we simplify the expression. Factor the numerator: $$4 - x^2 = (2 - x)(2 + x)$$ 4. Substitute the factorization back into the limit expression: $$\frac{(2 - x)(2 + x)}{2 + x}$$ 5. Cancel the common factor $(2 + x)$ (valid for $x \neq -2$): $$2 - x$$ 6. Now, find the limit by direct substitution of $x = -2$ in the simplified expression: $$2 - (-2) = 2 + 2 = 4$$ 7. Therefore, the limit is: $$\boxed{4}$$