1. The problem is to find the limit $$\lim_{x \to 7} \frac{x^2 - 4x - 21}{x^2 - 49}.$$\n\n2. Substitute $x = 7$ directly to check if the expression is defined:\n$$\frac{7^2 - 4(7) - 21}{7^2 - 49} = \frac{49 - 28 - 21}{49 - 49} = \frac{0}{0},$$\nwhich is indeterminate. So we need to simplify the expression.\n\n3. Factor numerator and denominator:\n- Numerator: $x^2 - 4x - 21 = (x - 7)(x + 3)$ because $-7 \times 3 = -21$ and $-7 + 3 = -4$.\n- Denominator: $x^2 - 49 = (x - 7)(x + 7)$ (difference of squares).\n\n4. Rewrite the limit expression with factors:\n$$\frac{(x - 7)(x + 3)}{(x - 7)(x + 7)}.$$\n\n5. Cancel the common factor $(x - 7)$ (which is valid since we are taking the limit as $x \to 7$, not at $x=7$):\n$$\frac{x + 3}{x + 7}.$$\n\n6. Now substitute $x = 7$ again:\n$$\frac{7 + 3}{7 + 7} = \frac{10}{14} = \frac{5}{7}.$$\n\nAnswer: $$\lim_{x \to 7} \frac{x^2 - 4x - 21}{x^2 - 49} = \frac{5}{7}.$$