1. **State the problem:** Find the limit $$\lim_{x \to 1} g(x)$$ where $$g(x) = \frac{x^3 - 1}{x - 1}$$. 2. **Recall the formula and rules:** The expression is a rational function that becomes indeterminate of the form $$\frac{0}{0}$$ at $$x=1$$. To find the limit, we simplify the expression. 3. **Factor the numerator:** Use the difference of cubes formula: $$x^3 - 1 = (x - 1)(x^2 + x + 1)$$. 4. **Simplify the expression:** $$g(x) = \frac{x^3 - 1}{x - 1} = \frac{\cancel{(x - 1)}(x^2 + x + 1)}{\cancel{(x - 1)}} = x^2 + x + 1$$. 5. **Evaluate the limit:** Substitute $$x = 1$$ into the simplified expression: $$\lim_{x \to 1} g(x) = 1^2 + 1 + 1 = 3$$. 6. **Interpretation:** As $$x$$ approaches 1, $$g(x)$$ approaches 3, which matches the behavior suggested by the table values approaching 2 from the left (likely a typo or different function in the table). The correct limit is 3. **Final answer:** $$\boxed{3}$$