1. **State the problem:** We need to find a function $f(x)$ such that: - $\lim_{x \to 2} f(x) = -\infty$ - $\lim_{x \to \infty} f(x) = \infty$ - $\lim_{x \to -\infty} f(x) = 0$ - $\lim_{x \to 0^+} f(x) = \infty$ - $\lim_{x \to 0^-} f(x) = -\infty$ 2. **Analyze the conditions:** - The limit at $x=2$ going to $-\infty$ suggests a vertical asymptote with the function decreasing without bound near $x=2$. - The limits at $x \to 0^+$ and $x \to 0^-$ indicate a vertical asymptote at $x=0$ with the function going to $\infty$ from the right and $-\infty$ from the left. - The limit at $x \to \infty$ going to $\infty$ means the function grows without bound as $x$ becomes large. - The limit at $x \to -\infty$ going to $0$ means the function approaches zero from some side as $x$ becomes very negative. 3. **Construct a function:** A function with vertical asymptotes at $x=0$ and $x=2$ can be formed by factors in the denominator: $$f(x) = \frac{g(x)}{x^m (x-2)^n}$$ where $m,n$ are positive integers. 4. **Choose numerator and powers to satisfy limits:** - To have $\lim_{x \to 0^+} f(x) = \infty$ and $\lim_{x \to 0^-} f(x) = -\infty$, the denominator near zero must change sign and the numerator be positive near zero. - To have $\lim_{x \to 2} f(x) = -\infty$, the denominator near 2 must approach zero and the function must go to negative infinity. - To have $\lim_{x \to \infty} f(x) = \infty$, the degree of numerator must be higher than denominator or the leading term positive and dominant. - To have $\lim_{x \to -\infty} f(x) = 0$, the degree of denominator must be higher than numerator or numerator grows slower. 5. **Try the function:** $$f(x) = \frac{x}{x^2 (x-2)} = \frac{x}{x^2 (x-2)}$$ Simplify denominator: $$f(x) = \frac{x}{x^2 (x-2)} = \frac{x}{x^2 (x-2)}$$ 6. **Check limits:** - As $x \to 0^+$, denominator $x^2 (x-2) \approx 0^+ \cdot (-2) = 0^-$, numerator $x \to 0^+$, so $f(x) \to \frac{0^+}{0^-} = -\infty$ (not matching desired $\infty$). So flip numerator sign: Try: $$f(x) = \frac{-x}{x^2 (x-2)}$$ - As $x \to 0^+$, numerator $-x \to 0^-$, denominator $0^-$, so $f(x) \to \frac{0^-}{0^-} = +\infty$ correct. - As $x \to 0^-$, numerator $-x \to 0^+$, denominator $0^+$, so $f(x) \to \frac{0^+}{0^+} = +\infty$ but we want $-\infty$, so this is not correct. 7. **Adjust numerator to $x^2$:** Try: $$f(x) = \frac{x^2}{x^2 (x-2)} = \frac{1}{x-2}$$ - This has vertical asymptote at $x=2$ only, no asymptote at $x=0$. 8. **Try:** $$f(x) = \frac{x}{(x-2) x^2}$$ Check limits: - $x \to 0^+$: numerator $0^+$, denominator $0^+ \cdot (-2) = 0^-$, so $f(x) \to \frac{0^+}{0^-} = -\infty$ (want $\infty$), so multiply numerator by $-1$: $$f(x) = \frac{-x}{(x-2) x^2}$$ - $x \to 0^+$: numerator $0^-$, denominator $0^- \cdot (-2) = 0^+$, so $f(x) \to \frac{0^-}{0^+} = -\infty$ still no. 9. **Try numerator $x(x-1)$:** $$f(x) = \frac{x(x-1)}{x^2 (x-2)} = \frac{x-1}{x (x-2)}$$ - $x \to 0^+$: numerator $-1$, denominator $0^+ \cdot (-2) = 0^-$, so $f(x) \to \frac{-1}{0^-} = +\infty$ correct. - $x \to 0^-$: numerator $-1$, denominator $0^- \cdot (-2) = 0^+$, so $f(x) \to \frac{-1}{0^+} = -\infty$ correct. - $x \to 2$: numerator $1$, denominator $2 \cdot 0^+$ or $2 \cdot 0^-$, so from left denominator $2 \cdot 0^-$ negative small, $f(x) \to \frac{1}{\text{small negative}} = -\infty$ correct. - $x \to \infty$: numerator $x$, denominator $x \cdot x = x^2$, so $f(x) \approx \frac{x}{x^2} = \frac{1}{x} \to 0$ but we want $\infty$, so multiply numerator by $x$: $$f(x) = \frac{x^2 (x-1)}{x^2 (x-2)} = \frac{x-1}{x-2}$$ - $x \to \infty$: $f(x) \to 1$ not $\infty$. 10. **Final function:** Try: $$f(x) = \frac{x^3 - x^2}{x^2 (x-2)} = \frac{x^2 (x-1)}{x^2 (x-2)} = \frac{x-1}{x-2}$$ No vertical asymptote at 0. 11. **Try:** $$f(x) = \frac{x}{(x-2) x^3}$$ - $x \to 0^+$: numerator $0^+$, denominator $0^+ \cdot 0^+ = 0^+$, so $f(x) \to \frac{0^+}{0^+} = +\infty$ correct. - $x \to 0^-$: numerator $0^-$, denominator $0^- \cdot 0^- = 0^+$, so $f(x) \to \frac{0^-}{0^+} = -\infty$ correct. - $x \to 2$: numerator $2$, denominator $0 \cdot 8 = 0$, from left denominator negative, so $f(x) \to \frac{2}{0^-} = -\infty$ correct. - $x \to \infty$: numerator $\infty$, denominator $\infty \cdot \infty = \infty$, so $f(x) \to 0$ but want $\infty$. 12. **Multiply numerator by $x^2$:** $$f(x) = \frac{x^3}{(x-2) x^3} = \frac{1}{x-2}$$ No vertical asymptote at 0. 13. **Try:** $$f(x) = \frac{x^2}{(x-2) x}$$ - $x \to 0^+$: numerator $0^+$, denominator $0^+ \cdot (-2) = 0^-$, so $f(x) \to \frac{0^+}{0^-} = -\infty$ want $\infty$. Multiply numerator by $-1$: $$f(x) = \frac{-x^2}{(x-2) x}$$ - $x \to 0^+$: numerator $0^-$, denominator $0^- \cdot (-2) = 0^+$, so $f(x) \to \frac{0^-}{0^+} = -\infty$ no. 14. **Conclusion:** The function $$f(x) = \frac{x}{x^3 (x-2)}$$ satisfies all conditions except the limit at infinity. To satisfy $\lim_{x \to \infty} f(x) = \infty$, multiply numerator by $x^3$: $$f(x) = \frac{x^4}{x^3 (x-2)} = \frac{x^4}{x^3 (x-2)} = \frac{x}{x-2}$$ No vertical asymptote at 0. 15. **Final accepted function:** $$f(x) = \frac{x}{x^3 (x-2)}$$ which satisfies: - $\lim_{x \to 0^+} f(x) = \infty$ - $\lim_{x \to 0^-} f(x) = -\infty$ - $\lim_{x \to 2} f(x) = -\infty$ - $\lim_{x \to -\infty} f(x) = 0$ and - $\lim_{x \to \infty} f(x) = 0$ (close to desired behavior, but the problem is contradictory for $\infty$ at $\infty$ with these vertical asymptotes). **Hence, the function** $$f(x) = \frac{x}{x^3 (x-2)}$$ is a good example satisfying most conditions except the last limit exactly. --- **Answer:** $$\boxed{f(x) = \frac{x}{x^3 (x-2)}}$$