1. **State the problem:** We have a piecewise function $$f(x) = \begin{cases} 4x^2 + ax - 1 & \text{if } 0 \leq x < 2 \\ 3x + a & \text{if } x > 2 \end{cases}$$ We want to find the values of $a$ and $L$ such that $$\lim_{x \to 2} f(x) = L$$ 2. **Recall the limit condition for piecewise functions:** For the limit at $x=2$ to exist, the left-hand limit and right-hand limit must be equal: $$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = L$$ 3. **Calculate the left-hand limit:** $$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (4x^2 + ax - 1) = 4(2)^2 + a(2) - 1 = 16 + 2a - 1 = 15 + 2a$$ 4. **Calculate the right-hand limit:** $$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3x + a) = 3(2) + a = 6 + a$$ 5. **Set the limits equal to find $a$:** $$15 + 2a = 6 + a$$ Subtract $a$ from both sides: $$15 + \cancel{2a} - \cancel{a} = 6 + \cancel{a} - \cancel{a} \implies 15 + a = 6$$ Subtract 15 from both sides: $$\cancel{15} + a - \cancel{15} = 6 - 15 \implies a = -9$$ 6. **Find $L$ by substituting $a = -9$ into either limit:** Using right-hand limit: $$L = 6 + a = 6 + (-9) = -3$$ **Final answers:** $$a = -9$$ $$L = -3$$