1. **State the problem:** We want to evaluate the limit $$\lim_{x \to 8} \left( \frac{1}{x-8} - \frac{16}{x^2-64} \right).$$ 2. **Rewrite the expression:** Notice that $$x^2 - 64 = (x-8)(x+8).$$ So the expression becomes $$\frac{1}{x-8} - \frac{16}{(x-8)(x+8)}.$$ 3. **Find a common denominator:** The common denominator is $$(x-8)(x+8).$$ Rewrite each term: $$\frac{1}{x-8} = \frac{x+8}{(x-8)(x+8)}$$ 4. **Combine the fractions:** $$\frac{x+8}{(x-8)(x+8)} - \frac{16}{(x-8)(x+8)} = \frac{x+8 - 16}{(x-8)(x+8)} = \frac{x - 8}{(x-8)(x+8)}.$$ 5. **Simplify the fraction:** $$\frac{\cancel{x - 8}}{\cancel{x - 8}(x+8)} = \frac{1}{x+8}.$$ 6. **Rewrite the limit:** $$\lim_{x \to 8} \left( \frac{1}{x-8} - \frac{16}{x^2-64} \right) = \lim_{x \to 8} \frac{1}{x+8}.$$ 7. **Evaluate the limit by direct substitution:** $$\frac{1}{8+8} = \frac{1}{16}.$$ **Final answer:** $$\lim_{x \to 8} \left( \frac{1}{x-8} - \frac{16}{x^2-64} \right) = \frac{1}{16}.$$