1. **State the problem:** Find the limit $$\lim_{x \to 2} \frac{x^3 - 8}{3 - \sqrt{2x + 5}}$$. 2. **Recognize the indeterminate form:** Substitute $x=2$ directly: $$\frac{2^3 - 8}{3 - \sqrt{2(2) + 5}} = \frac{8 - 8}{3 - \sqrt{4 + 5}} = \frac{0}{3 - 3} = \frac{0}{0}$$ which is indeterminate. 3. **Use algebraic manipulation:** Factor the numerator using difference of cubes: $$x^3 - 8 = (x - 2)(x^2 + 2x + 4)$$ 4. **Rationalize the denominator:** Multiply numerator and denominator by the conjugate of the denominator: $$\frac{(x - 2)(x^2 + 2x + 4)}{3 - \sqrt{2x + 5}} \times \frac{3 + \sqrt{2x + 5}}{3 + \sqrt{2x + 5}} = \frac{(x - 2)(x^2 + 2x + 4)(3 + \sqrt{2x + 5})}{(3)^2 - (\sqrt{2x + 5})^2}$$ 5. **Simplify the denominator:** $$9 - (2x + 5) = 9 - 2x - 5 = 4 - 2x = 2(2 - x)$$ 6. **Rewrite the expression:** $$\frac{(x - 2)(x^2 + 2x + 4)(3 + \sqrt{2x + 5})}{2(2 - x)}$$ 7. **Cancel common factors:** Note that $2 - x = -(x - 2)$, so: $$\frac{(x - 2)(x^2 + 2x + 4)(3 + \sqrt{2x + 5})}{2 \cancel{(2 - x)}} = \frac{(x - 2)(x^2 + 2x + 4)(3 + \sqrt{2x + 5})}{2 \cancel{-(x - 2)}} = -\frac{(x^2 + 2x + 4)(3 + \sqrt{2x + 5})}{2}$$ 8. **Evaluate the limit by substituting $x=2$:** $$-\frac{(2^2 + 2 \times 2 + 4)(3 + \sqrt{2 \times 2 + 5})}{2} = -\frac{(4 + 4 + 4)(3 + \sqrt{9})}{2} = -\frac{12 \times (3 + 3)}{2} = -\frac{12 \times 6}{2} = -\frac{72}{2} = -36$$ **Final answer:** $$\boxed{-36}$$