1. **State the problem:** Evaluate the limits $$\lim_{x \to 7^+} \frac{\sqrt{x-7}}{\sqrt{x^2 - 49}}, \quad \lim_{x \to 7^-} \frac{\sqrt{x-7}}{\sqrt{x^2 - 49}}, \quad \lim_{x \to 7} \frac{\sqrt{x-7}}{\sqrt{x^2 - 49}}$$ 2. **Rewrite the expression:** Note that $x^2 - 49 = (x-7)(x+7)$, so $$\frac{\sqrt{x-7}}{\sqrt{x^2 - 49}} = \frac{\sqrt{x-7}}{\sqrt{(x-7)(x+7)}} = \frac{\sqrt{x-7}}{\sqrt{x-7} \sqrt{x+7}}$$ 3. **Simplify the expression:** For $x \neq 7$, we can cancel $\sqrt{x-7}$: $$\frac{\sqrt{x-7}}{\cancel{\sqrt{x-7}} \sqrt{x+7}} = \frac{1}{\sqrt{x+7}}$$ 4. **Consider the domain:** The expression $\sqrt{x-7}$ is defined only for $x \geq 7$. For $x < 7$, $\sqrt{x-7}$ is not real, so the limit from the left does not exist in the real numbers. 5. **Evaluate the right-hand limit:** $$\lim_{x \to 7^+} \frac{1}{\sqrt{x+7}} = \frac{1}{\sqrt{7+7}} = \frac{1}{\sqrt{14}}$$ 6. **Evaluate the left-hand limit:** Since $x < 7$ implies $x-7 < 0$, $\sqrt{x-7}$ is not real, so $$\lim_{x \to 7^-} \frac{\sqrt{x-7}}{\sqrt{x^2 - 49}} \text{ does not exist (in real numbers)}$$ 7. **Evaluate the two-sided limit:** Since the left-hand limit does not exist, the two-sided limit $$\lim_{x \to 7} \frac{\sqrt{x-7}}{\sqrt{x^2 - 49}}$$ also does not exist. **Final answers:** $$\lim_{x \to 7^+} \frac{\sqrt{x-7}}{\sqrt{x^2 - 49}} = \frac{1}{\sqrt{14}}$$ $$\lim_{x \to 7^-} \frac{\sqrt{x-7}}{\sqrt{x^2 - 49}} \text{ does not exist}$$ $$\lim_{x \to 7} \frac{\sqrt{x-7}}{\sqrt{x^2 - 49}} \text{ does not exist}$$