1. **State the problem:** Evaluate the limit $$\lim_{x \to 6} \frac{x^2 - 32}{2x - 5}$$. 2. **Substitute the value:** First, substitute $x=6$ into numerator and denominator: $$\text{numerator} = 6^2 - 32 = 36 - 32 = 4$$ $$\text{denominator} = 2(6) - 5 = 12 - 5 = 7$$ 3. **Check if direct substitution is valid:** Since denominator is not zero, the limit is simply the fraction value at $x=6$: $$\lim_{x \to 6} \frac{x^2 - 32}{2x - 5} = \frac{4}{7}$$ **Final answer:** $$\boxed{\frac{4}{7}}$$