1. **State the problem:** Find the limit $$\lim_{x \to 1} \frac{x-1}{x^2 + x - 2}$$. 2. **Recall the formula and rules:** The limit of a quotient is the quotient of the limits, provided the denominator's limit is not zero. That is, if $$\lim_{x \to p} f(x) = L_1$$ and $$\lim_{x \to p} g(x) = L_2 \neq 0$$, then $$\lim_{x \to p} \frac{f(x)}{g(x)} = \frac{L_1}{L_2}$$. 3. **Evaluate the denominator at $$x=1$$:** $$x^2 + x - 2 = 1^2 + 1 - 2 = 1 + 1 - 2 = 0$$ Since the denominator is zero at $$x=1$$, direct substitution is not possible. 4. **Factor the denominator:** $$x^2 + x - 2 = (x - 1)(x + 2)$$ 5. **Rewrite the expression:** $$\frac{x-1}{(x-1)(x+2)}$$ 6. **Simplify by canceling common factors:** For $$x \neq 1$$, $$\frac{x-1}{(x-1)(x+2)} = \frac{1}{x+2}$$ 7. **Find the limit of the simplified expression as $$x \to 1$$:** $$\lim_{x \to 1} \frac{1}{x+2} = \frac{1}{1+2} = \frac{1}{3}$$ **Final answer:** $$\boxed{\frac{1}{3}}$$