1. We are asked to find the limit: $$\lim_{x \to 2^+} \frac{x + 2}{x^2 - 4}.$$ 2. The denominator can be factored using the difference of squares formula: $$x^2 - 4 = (x - 2)(x + 2).$$ 3. Substitute this into the expression: $$\frac{x + 2}{(x - 2)(x + 2)}.$$ 4. We can cancel the common factor $x + 2$ in numerator and denominator: $$\frac{\cancel{x + 2}}{(x - 2)\cancel{(x + 2)}} = \frac{1}{x - 2}.$$ 5. Now evaluate the limit as $x$ approaches $2$ from the right: $$\lim_{x \to 2^+} \frac{1}{x - 2}.$$ 6. As $x$ approaches $2$ from the right, $x - 2$ is a very small positive number, so $\frac{1}{x - 2}$ becomes very large positive. 7. Therefore, the limit is $+\infty$. **Final answer:** D. $\infty$