1. **Problem:** Find the limit $$\lim_{x \to 1} \frac{x^3 - 3x}{x - 1}$$ using the values $$x = \{0.9, 0.99, 0.999, 1.001, 1.01, 1.1\}$$. 2. **Formula and rules:** The limit of a function as $$x$$ approaches a value is the value that $$f(x)$$ approaches as $$x$$ gets closer to that value. 3. **Simplify the function:** Factor the numerator: $$x^3 - 3x = x(x^2 - 3)$$ So, $$f(x) = \frac{x(x^2 - 3)}{x - 1}$$ 4. **Evaluate numerically:** Calculate $$f(x)$$ for each given $$x$$: - $$f(0.9) = \frac{0.9(0.9^2 - 3)}{0.9 - 1} = \frac{0.9(0.81 - 3)}{-0.1} = \frac{0.9(-2.19)}{-0.1} = 19.71$$ - $$f(0.99) = \frac{0.99(0.9801 - 3)}{-0.01} = \frac{0.99(-2.0199)}{-0.01} = 200.9701$$ - $$f(0.999) = \frac{0.999(0.998001 - 3)}{-0.001} = \frac{0.999(-2.001999)}{-0.001} = 2000.997$$ - $$f(1.001) = \frac{1.001(1.002001 - 3)}{0.001} = \frac{1.001(-1.997999)}{0.001} = -2000.997$$ - $$f(1.01) = \frac{1.01(1.0201 - 3)}{0.01} = \frac{1.01(-1.9799)}{0.01} = -199.9799$$ - $$f(1.1) = \frac{1.1(1.21 - 3)}{0.1} = \frac{1.1(-1.79)}{0.1} = -19.69$$ 5. **Interpretation:** As $$x$$ approaches 1 from the left, $$f(x)$$ grows very large positive; from the right, it grows very large negative. So, the limit does not exist (it diverges). **Final answer:** $$\lim_{x \to 1} \frac{x^3 - 3x}{x - 1}$$ does not exist because the left and right limits are not equal.