1. The problem is to find the limit: $$\lim_{x \to 3} \frac{x^2 + x - 12}{\sqrt{5x + 1} - 4}$$. 2. First, substitute $x = 3$ directly to check if the expression is defined: $$\frac{3^2 + 3 - 12}{\sqrt{5 \cdot 3 + 1} - 4} = \frac{9 + 3 - 12}{\sqrt{15 + 1} - 4} = \frac{0}{4 - 4} = \frac{0}{0}$$ which is an indeterminate form. 3. To resolve this, we use algebraic manipulation. The numerator factors as: $$x^2 + x - 12 = (x + 4)(x - 3)$$ 4. The denominator has a square root difference. Multiply numerator and denominator by the conjugate of the denominator: $$\frac{(x + 4)(x - 3)}{\sqrt{5x + 1} - 4} \times \frac{\sqrt{5x + 1} + 4}{\sqrt{5x + 1} + 4} = \frac{(x + 4)(x - 3)(\sqrt{5x + 1} + 4)}{(\sqrt{5x + 1})^2 - 4^2}$$ 5. Simplify the denominator: $$ (\sqrt{5x + 1})^2 - 16 = 5x + 1 - 16 = 5x - 15 = 5(x - 3)$$ 6. Now the expression is: $$\frac{(x + 4)(x - 3)(\sqrt{5x + 1} + 4)}{5(x - 3)}$$ 7. Cancel the common factor $(x - 3)$: $$\frac{(x + 4)(\sqrt{5x + 1} + 4)}{5}$$ 8. Substitute $x = 3$ now: $$\frac{(3 + 4)(\sqrt{5 \cdot 3 + 1} + 4)}{5} = \frac{7(\sqrt{16} + 4)}{5} = \frac{7(4 + 4)}{5} = \frac{7 \times 8}{5} = \frac{56}{5} = 11.2$$ Final answer: $$\lim_{x \to 3} \frac{x^2 + x - 12}{\sqrt{5x + 1} - 4} = \frac{56}{5} = 11.2$$