Limit Rational 43D897

1. **State the problem:** Find the limit $$\lim_{x \to 1} \frac{x^2 + 4x - 5}{x^2 - 1}$$. 2. **Check direct substitution:** Substitute $x=1$ into numerator and denominator: $$1^2 + 4(1) - 5 = 1 + 4 - 5 = 0$$ $$1^2 - 1 = 1 - 1 = 0$$ Both numerator and denominator equal zero, so direct substitution gives an indeterminate form $\frac{0}{0}$. 3. **Factor numerator and denominator:** $$x^2 + 4x - 5 = (x + 5)(x - 1)$$ $$x^2 - 1 = (x - 1)(x + 1)$$ 4. **Simplify the expression by canceling common factors:** $$\frac{(x + 5)\cancel{(x - 1)}}{\cancel{(x - 1)}(x + 1)} = \frac{x + 5}{x + 1}$$ 5. **Evaluate the simplified limit:** Substitute $x=1$: $$\frac{1 + 5}{1 + 1} = \frac{6}{2} = 3$$ **Final answer:** $$\lim_{x \to 1} \frac{x^2 + 4x - 5}{x^2 - 1} = 3$$