1. **Problem statement:** Given the function $f(x) = \frac{x^3 - 4x^2 + 1}{x - 2}$, find its domain, simplify the expression if possible, determine the nature of the discontinuity at $x=2$, and compute $\lim_{x \to 2} f(x)$. 2. **Domain:** The function is a rational function and is undefined where the denominator is zero. So, set denominator equal to zero: $$x - 2 = 0 \implies x = 2$$ Thus, the domain is all real numbers except $x=2$. 3. **Simplify the expression:** Try to factor the numerator to see if $(x-2)$ is a factor: Use polynomial division or factor by synthetic division: Divide $x^3 - 4x^2 + 1$ by $x - 2$. Perform polynomial division: $$\frac{x^3 - 4x^2 + 1}{x - 2} = x^2 - 2x - 4 + \frac{\cancel{9}}{x - 2}$$ Since the remainder is 9, $(x-2)$ is not a factor, so no simplification by cancellation. 4. **Determine the discontinuity at $x=2$:** Since the denominator is zero at $x=2$ and the numerator is not zero at $x=2$ (evaluate numerator at 2: $2^3 - 4(2)^2 + 1 = 8 - 16 + 1 = -7 \neq 0$), the function has a vertical asymptote at $x=2$. 5. **Compute the limit as $x$ approaches 2:** Calculate: $$\lim_{x \to 2} \frac{x^3 - 4x^2 + 1}{x - 2}$$ Since direct substitution gives division by zero, analyze the limit from left and right: - Numerator at $x=2$ is $-7$ (negative). - Denominator approaches 0 from positive side when $x \to 2^+$ and from negative side when $x \to 2^-$. Therefore: $$\lim_{x \to 2^-} f(x) = \frac{-7}{0^-} = +\infty$$ $$\lim_{x \to 2^+} f(x) = \frac{-7}{0^+} = -\infty$$ The limit does not exist because the left and right limits are not equal. **Final answers:** - Domain: $\{x \in \mathbb{R} : x \neq 2\}$ - Simplified form: No simplification possible - Discontinuity at $x=2$: Vertical asymptote - Limit as $x \to 2$: Does not exist (left limit $+\infty$, right limit $-\infty$)