## Problem Let $f(x)=\frac{x^3-4x^2+1}{x-2}$. Find the domain of $f$, simplify where possible, determine whether $x=2$ is removable or a vertical asymptote, and compute $\lim_{x\to 2} f(x)$. 1. **List the domain restrictions** - A rational function is undefined where the denominator is $0$. - Denominator: $x-2$. - So the domain excludes $x=2$. $$\text{Domain: }\{x\mid x\neq 2\}$$ 2. **Factor/simplify to see what happens near $x=2$** - Check whether the numerator has a factor of $x-2$. - Substitute $x=2$ into the numerator: $$2^3-4\cdot 2^2+1=8-16+1=-7$$ - Since the result is not $0$, $(x-2)$ is **not** a factor of the numerator. - Therefore, no cancellation is possible. - Keep the expression as-is: $$f(x)=\frac{x^3-4x^2+1}{x-2},\quad x\neq 2$$ 3. **Classify the discontinuity at $x=2$** - Because $(x-2)$ does not cancel, the function has a blow-up at $x=2$. $$x=2\text{ is a vertical asymptote (not removable)}$$ 4. **Compute $\lim_{x\to 2} f(x)$** - Use the fact that for a rational function with no cancellation, the limit is infinite in magnitude if the numerator is nonzero. - Evaluate the numerator at $x=2$: $$x^3-4x^2+1\Big\vert_{x=2}=-7$$ - Compare the signs near $x=2$: - If $x\to 2^+$ then $x-2>0$, so $\frac{-7}{x-2}\to -\infty$. - If $x\to 2^-$ then $x-2<0$, so $\frac{-7}{x-2}\to +\infty$. $$\lim_{x\to 2^-} f(x)=+\infty$$ $$\lim_{x\to 2^+} f(x)=-\infty$$ - Since the one-sided limits are not equal, the two-sided limit does not exist (it diverges). $$\lim_{x\to 2} f(x)\text{ does not exist (diverges)}$$ ## Final Answers 1. Domain: $\{x\mid x\neq 2\}$. 2. Simplified form: $f(x)=\frac{x^3-4x^2+1}{x-2}$ (no cancellation). 3. $x=2$ is a vertical asymptote. 4. $\lim_{x\to 2} f(x)$ diverges: $+\infty$ from the left and $-\infty$ from the right.