1. **State the problem:** Find the limit $$\lim_{x \to 2} k(x)$$ where $$k(x) = \frac{x^3 - 2x - 4}{x - 2}$$. 2. **Understand the problem:** Direct substitution of $$x=2$$ gives $$\frac{8 - 4 - 4}{0} = \frac{0}{0}$$ which is indeterminate. We need to simplify the expression. 3. **Factor the numerator:** We try to factor $$x^3 - 2x - 4$$. 4. **Use polynomial division or factor theorem:** Since $$x=2$$ makes numerator zero, $$x-2$$ is a factor. Divide $$x^3 - 2x - 4$$ by $$x-2$$: $$\begin{aligned} &x^3 - 0x^2 - 2x - 4 \div (x - 2) \\ &= x^2 + 2x + 2 \end{aligned}$$ 5. **Rewrite the function:** $$k(x) = \frac{(x-2)(x^2 + 2x + 2)}{x-2}$$ 6. **Cancel common factors:** $$k(x) = \cancel{\frac{(x-2)}{(x-2)}} (x^2 + 2x + 2) = x^2 + 2x + 2, \quad x \neq 2$$ 7. **Evaluate the limit:** $$\lim_{x \to 2} k(x) = \lim_{x \to 2} (x^2 + 2x + 2) = 2^2 + 2 \times 2 + 2 = 4 + 4 + 2 = 10$$ **Final answer:** $$\boxed{10}$$