1. **State the problem:** Find the limit $$\lim_{x\to 5}\frac{x^2-25}{x^2-5x}$$. 2. **Recall the formula and rules:** When direct substitution results in an indeterminate form like $$\frac{0}{0}$$, we try to simplify the expression by factoring and canceling common factors. 3. **Substitute directly:** $$\frac{5^2-25}{5^2-5\cdot 5} = \frac{25-25}{25-25} = \frac{0}{0}$$ which is indeterminate. 4. **Factor numerator and denominator:** $$\frac{x^2-25}{x^2-5x} = \frac{(x-5)(x+5)}{x(x-5)}$$ 5. **Cancel common factor:** $$\frac{\cancel{(x-5)}(x+5)}{x\cancel{(x-5)}} = \frac{x+5}{x}$$ 6. **Evaluate the simplified limit:** $$\lim_{x\to 5} \frac{x+5}{x} = \frac{5+5}{5} = \frac{10}{5} = 2$$ **Final answer:** $$2$$