1. Stating the problem: Calculate the left-hand limit as $x$ approaches 1 from the left for the function $$\frac{2x + 3}{x^2 - 1}$$. 2. Recall the formula and rules: The limit $$\lim_{x \to a^-} f(x)$$ means we approach $a$ from values less than $a$. 3. Simplify the denominator: $$x^2 - 1 = (x - 1)(x + 1)$$. 4. Substitute values close to 1 from the left (e.g., $x = 0.9$): - Numerator: $2(0.9) + 3 = 1.8 + 3 = 4.8$ - Denominator: $(0.9 - 1)(0.9 + 1) = (-0.1)(1.9) = -0.19$ 5. The fraction near $x=1^-$ is approximately $$\frac{4.8}{-0.19} = -25.26$$, a large negative number. 6. As $x$ approaches 1 from the left, the numerator approaches $2(1) + 3 = 5$ (positive), and the denominator approaches $(1 - 1)(1 + 1) = 0 \times 2 = 0$. 7. Since $(x - 1)$ is negative just before 1, and $(x + 1)$ is positive, the denominator approaches 0 through negative values. 8. Therefore, the fraction approaches $$\frac{5}{0^-} = -\infty$$. Final answer: $$\lim_{x \to 1^-} \frac{2x + 3}{x^2 - 1} = -\infty$$.