1. First, state the problem: Find the limit as $x$ approaches 1 of the function $$\frac{3x^3 + 2x^2 - 3x}{x^3 - 1}$$. 2. Substitute $x = 1$ directly into the expression to check if it's defined: $$\frac{3(1)^3 + 2(1)^2 - 3(1)}{(1)^3 - 1} = \frac{3 + 2 - 3}{1 - 1} = \frac{2}{0}$$ The denominator is zero, so the expression is undefined at $x=1$. 3. Since direct substitution leads to division by zero, try to factor numerator and denominator to simplify: Denominator: $$x^3 - 1 = (x - 1)(x^2 + x + 1)$$ 4. Factor the numerator: $$3x^3 + 2x^2 - 3x = x(3x^2 + 2x - 3)$$ Now factor the quadratic $3x^2 + 2x - 3$: Find two numbers $a,b$ such that $a\times b = 3 \times (-3) = -9$ and $a + b = 2$. The numbers are 3 and -1. Rewrite quadratic: $$3x^2 + 3x - x - 3 = 3x(x + 1) - 1(x + 1) = (3x - 1)(x + 1)$$ 5. So numerator becomes: $$x(3x - 1)(x + 1)$$ 6. The original expression can now be written as: $$\frac{x(3x - 1)(x + 1)}{(x - 1)(x^2 + x + 1)}$$ 7. Since $x-1$ is not a factor of numerator, the expression cannot be simplified further. 8. Evaluate the numerator and denominator when $x \to 1$: Numerator: $$1 \times (3(1) - 1) \times (1 + 1) = 1 \times 2 \times 2 = 4$$ Denominator approaches zero ($1 - 1 = 0$). 9. Evaluate the limit from the left and right of 1 to determine limit behavior: - When $x\to 1^-$, denominator $x-1 < 0$, numerator near 4 positive, so total fraction $\to -\infty$. - When $x\to 1^+$, denominator $x-1 > 0$, numerator near 4 positive, so fraction $\to +\infty$. 10. Hence, the limit does not exist (it diverges to $-\infty$ from the left and $+\infty$ from the right). **Final answer:** $$\lim_{x \to 1} \frac{3x^3 + 2x^2 - 3x}{x^3 - 1}$$ does not exist because the left and right limits are infinite with opposite signs.