1. **State the problem:** We need to find the limits of the function $$f(x) = \frac{1 - 2x}{x^2 - 1}$$ as $$x$$ approaches $$-1$$ from the left ($$x \to -1^-$$), from the right ($$x \to -1^+$$), and the overall limit $$x \to -1$$. 2. **Factor the denominator:** The denominator can be factored as $$x^2 - 1 = (x-1)(x+1)$$. 3. **Rewrite the function:** $$f(x) = \frac{1 - 2x}{(x-1)(x+1)}$$ 4. **Evaluate the behavior as $$x \to -1$$:** Substitute $$x = -1$$ in numerator: $$1 - 2(-1) = 1 + 2 = 3$$ Substitute $$x = -1$$ in denominator: $$(x-1)(x+1) = (-1 - 1)(-1 + 1) = (-2)(0) = 0$$ The denominator approaches zero, numerator approaches 3, so the function approaches infinity or negative infinity. 5. **Check the sign of the denominator near $$x = -1$$:** - For $$x \to -1^-$$ (values slightly less than -1): - $$x-1$$ is slightly less than $$-2$$ (negative) - $$x+1$$ is slightly less than $$0$$ (negative) - Product of two negatives is positive - For $$x \to -1^+$$ (values slightly greater than -1): - $$x-1$$ is slightly less than $$0$$ (negative) - $$x+1$$ is slightly greater than $$0$$ (positive) - Product of negative and positive is negative 6. **Calculate limits:** (a) $$\lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} \frac{3}{\text{positive small}} = +\infty$$ (b) $$\lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} \frac{3}{\text{negative small}} = -\infty$$ (c) Since the left- and right-hand limits are not equal, $$\lim_{x \to -1} f(x)$$ does not exist. **Final answers:** $$\lim_{x \to -1^-} f(x) = +\infty$$ $$\lim_{x \to -1^+} f(x) = -\infty$$ $$\lim_{x \to -1} f(x) \text{ does not exist}$$