1. **State the problem:** Find the limit $$\lim_{x \to -1} \frac{2x^2 - x - 3}{x + 1}$$. 2. **Recall the limit rule:** If direct substitution leads to an indeterminate form like $$\frac{0}{0}$$, try to simplify the expression by factoring and canceling common factors. 3. **Evaluate the numerator at $$x = -1$$:** $$2(-1)^2 - (-1) - 3 = 2(1) + 1 - 3 = 2 + 1 - 3 = 0$$ 4. **Evaluate the denominator at $$x = -1$$:** $$-1 + 1 = 0$$ Since direct substitution gives $$\frac{0}{0}$$, simplify the expression. 5. **Factor the numerator:** $$2x^2 - x - 3 = (2x + 3)(x - 1)$$ 6. **Rewrite the limit:** $$\lim_{x \to -1} \frac{(2x + 3)(x - 1)}{x + 1}$$ 7. **Check for common factors:** There is no $$x + 1$$ factor in the numerator, so no cancellation is possible. 8. **Since no cancellation, direct substitution is not possible, but the denominator approaches zero. Let's check the behavior from left and right:** - For $$x \to -1^-$$, denominator $$x + 1 < 0$$ - For $$x \to -1^+$$, denominator $$x + 1 > 0$$ 9. **Evaluate numerator at $$x = -1$$:** 0, but near $$-1$$: - Numerator near $$-1$$: $$2x^2 - x - 3$$ - At $$x = -1.01$$: $$2(1.0201) - (-1.01) - 3 = 2.0402 + 1.01 - 3 = 0.0502 > 0$$ - At $$x = -0.99$$: $$2(0.9801) - (-0.99) - 3 = 1.9602 + 0.99 - 3 = -0.0498 < 0$$ 10. **So near $$-1$$ numerator changes sign, denominator changes sign, the limit does not exist because left and right limits differ.** **Final answer:** The limit $$\lim_{x \to -1} \frac{2x^2 - x - 3}{x + 1}$$ does not exist.