1. **Problem statement:** Find the limit $$\lim_{z \to 8} \frac{2z^2 - 17z + 8}{8 - z}$$. 2. **Formula and rules:** To find limits involving rational functions, first try direct substitution. If it results in an indeterminate form like $$\frac{0}{0}$$, factor and simplify the expression. 3. **Direct substitution:** Substitute $$z = 8$$: $$\frac{2(8)^2 - 17(8) + 8}{8 - 8} = \frac{2 \times 64 - 136 + 8}{0} = \frac{128 - 136 + 8}{0} = \frac{0}{0}$$ which is indeterminate. 4. **Factor numerator:** $$2z^2 - 17z + 8 = 2z^2 - 16z - z + 8 = 2z(z - 8) - 1(z - 8) = (2z - 1)(z - 8)$$. 5. **Rewrite the limit:** $$\lim_{z \to 8} \frac{(2z - 1)(z - 8)}{8 - z}$$. 6. **Simplify denominator:** Note that $$8 - z = -(z - 8)$$, so $$\frac{(2z - 1)(z - 8)}{8 - z} = \frac{(2z - 1)(z - 8)}{-(z - 8)} = -(2z - 1)$$ for $$z \neq 8$$. 7. **Evaluate simplified limit:** $$\lim_{z \to 8} -(2z - 1) = -(2 \times 8 - 1) = -(16 - 1) = -15$$. **Final answer:** $$\boxed{-15}$$