Question:

Step 1: Factor the denominator of the second fraction. We need to simplify the denominator $x^3 - 3x^2 + 2x$. First, factor out $x$: $$x^3 - 3x^2 + 2x = x(x^2 - 3x + 2)$$ Now factor the quadratic $x^2 - 3x + 2$: $$x^2 - 3x + 2 = (x-1)(x-2)$$ Therefore: $$x^3 - 3x^2 + 2x = x(x-1)(x-2)$$ Step 2: Rewrite the original expression with the factored denominator. The limit becomes: $$\lim_{x \to 2} \left[\frac{1}{x-2} - \frac{2(2x-3)}{x(x-1)(x-2)}\right]$$ Step 3: Combine the fractions over a common denominator. The common denominator is $x(x-1)(x-2)$. Rewrite the first fraction: $$\frac{1}{x-2} = \frac{x(x-1)}{x(x-1)(x-2)}$$ Now combine: $$\frac{x(x-1) - 2(2x-3)}{x(x-1)(x-2)}$$ Step 4: Simplify the numerator. Expand the numerator: $$x(x-1) - 2(2x-3) = x^2 - x - 4x + 6 = x^2 - 5x + 6$$ Factor this quadratic: $$x^2 - 5x + 6 = (x-2)(x-3)$$ Step 5: Cancel the common factor and simplify. The expression becomes: $$\frac{(x-2)(x-3)}{x(x-1)(x-2)} = \frac{x-3}{x(x-1)}$$ The factor $(x-2)$ cancels out, and the expression is now defined at $x = 2$. Step 6: Evaluate the limit. Substitute $x = 2$ into the simplified expression: $$\lim_{x \to 2} \frac{x-3}{x(x-1)} = \frac{2-3}{2(2-1)} = \frac{-1}{2 \cdot 1} = -\frac{1}{2}$$ **Final Answer:** The value of the limit is $-\dfrac{1}{2}$, which corresponds to **Option 3**.