1. We are asked to find the limit: $$\lim_{x \to 3} \frac{2x^2 - 5x - 3}{x-3}$$ 2. The formula for limits involving rational functions is to first try direct substitution. If direct substitution results in an indeterminate form like $$\frac{0}{0}$$, then we simplify the expression. 3. Substitute $x=3$ directly: $$\frac{2(3)^2 - 5(3) - 3}{3-3} = \frac{18 - 15 - 3}{0} = \frac{0}{0}$$ which is indeterminate. 4. Factor the numerator: $$2x^2 - 5x - 3 = (2x + 1)(x - 3)$$ 5. Rewrite the limit expression: $$\lim_{x \to 3} \frac{(2x + 1)(x - 3)}{x - 3}$$ 6. Cancel the common factor $x - 3$: $$\lim_{x \to 3} \frac{\cancel{(x - 3)}(2x + 1)}{\cancel{x - 3}} = \lim_{x \to 3} (2x + 1)$$ 7. Now substitute $x=3$: $$2(3) + 1 = 6 + 1 = 7$$ 8. Therefore, the limit is: $$\boxed{7}$$