1. **State the problem:** We have the function $$f(x) = \frac{x^3 - 4x^2 + 1}{x - 2}$$. We need to find its domain, simplify it if possible, determine the nature of the discontinuity at $$x=2$$, and compute $$\lim_{x \to 2} f(x)$$. 2. **Find the domain:** The denominator $$x - 2$$ cannot be zero, so $$x \neq 2$$. Thus, the domain is all real numbers except $$2$$. 3. **Simplify the expression:** We try to factor the numerator $$x^3 - 4x^2 + 1$$ to see if $$x-2$$ is a factor. Use polynomial division or synthetic division: Divide $$x^3 - 4x^2 + 0x + 1$$ by $$x - 2$$: - Leading term: $$x^3 \div x = x^2$$ - Multiply: $$x^2(x - 2) = x^3 - 2x^2$$ - Subtract: $$(x^3 - 4x^2) - (x^3 - 2x^2) = -2x^2$$ - Bring down next term: $$0x$$ - Divide: $$-2x^2 \div x = -2x$$ - Multiply: $$-2x(x - 2) = -2x^2 + 4x$$ - Subtract: $$(-2x^2 + 0x) - (-2x^2 + 4x) = -4x$$ - Bring down next term: $$+1$$ - Divide: $$-4x \div x = -4$$ - Multiply: $$-4(x - 2) = -4x + 8$$ - Subtract: $$( -4x + 1) - (-4x + 8) = 1 - 8 = -7$$ Remainder is $$-7$$, so $$x-2$$ is not a factor. 4. **Conclusion on simplification:** Since $$x-2$$ is not a factor, the expression cannot be simplified by canceling the denominator. 5. **Determine the discontinuity at $$x=2$$:** Since the denominator is zero at $$x=2$$ and the numerator is not zero (substitute $$x=2$$ into numerator: $$2^3 - 4(2)^2 + 1 = 8 - 16 + 1 = -7 \neq 0$$), the function has a vertical asymptote at $$x=2$$. 6. **Compute the limit as $$x$$ approaches 2:** $$\lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{x^3 - 4x^2 + 1}{x - 2}$$ Since the denominator approaches 0 and numerator approaches $$-7$$, the limit tends to $$\pm \infty$$ depending on the direction. - As $$x \to 2^+$$, denominator $$x-2 > 0$$, numerator $$\approx -7$$ (negative), so $$f(x) \to -\infty$$. - As $$x \to 2^-$$, denominator $$x-2 < 0$$, numerator $$\approx -7$$ (negative), so $$f(x) \to +\infty$$. Therefore, the limit does not exist (infinite discontinuity). **Final answers:** - Domain: $$\{x \in \mathbb{R} : x \neq 2\}$$ - Simplified form: cannot simplify - Discontinuity at $$x=2$$ is a vertical asymptote - $$\lim_{x \to 2} f(x)$$ does not exist (infinite limit)