1. We are asked to find the limit $ \lim_{x \to 1} \frac{x^3 - 2x + 1}{2x^3 + 3x - 5} $. 2. First, substitute $ x = 1 $ directly to check if the expression is defined: $$\frac{1^3 - 2(1) + 1}{2(1)^3 + 3(1) - 5} = \frac{1 - 2 + 1}{2 + 3 - 5} = \frac{0}{0}$$ This is an indeterminate form, so we need to simplify. 3. Factor numerator and denominator: Numerator: $ x^3 - 2x + 1 $. Try to factor by grouping or find roots. Check if $ x=1 $ is a root: $$1^3 - 2(1) + 1 = 0$$ So $ x-1 $ is a factor. Use polynomial division or synthetic division: Divide $ x^3 - 2x + 1 $ by $ x-1 $: $ x^3 - 2x + 1 = (x-1)(x^2 + x - 1) $. 4. Factor denominator $ 2x^3 + 3x - 5 $ similarly. Check if $ x=1 $ is a root: $$2(1)^3 + 3(1) - 5 = 2 + 3 - 5 = 0$$ So $ x-1 $ is a factor. Divide $ 2x^3 + 3x - 5 $ by $ x-1 $: Using synthetic division: Coefficients: 2, 0, 3, -5 Bring down 2, multiply by 1: 2, add to 0: 2 Multiply by 1: 2, add to 3: 5 Multiply by 1: 5, add to -5: 0 So quotient is $ 2x^2 + 2x + 5 $. 5. Rewrite the limit: $$\lim_{x \to 1} \frac{(x-1)(x^2 + x - 1)}{(x-1)(2x^2 + 2x + 5)}$$ 6. Cancel common factor $ x-1 $: $$\lim_{x \to 1} \frac{\cancel{(x-1)}(x^2 + x - 1)}{\cancel{(x-1)}(2x^2 + 2x + 5)} = \lim_{x \to 1} \frac{x^2 + x - 1}{2x^2 + 2x + 5}$$ 7. Substitute $ x = 1 $: Numerator: $1^2 + 1 - 1 = 1 + 1 - 1 = 1$ Denominator: $2(1)^2 + 2(1) + 5 = 2 + 2 + 5 = 9$ 8. Final answer: $$\boxed{\frac{1}{9}}$$