1. The problem states that the function $f(x)$ has horizontal asymptotes at $y = -4$ and $y = 6$, and a vertical asymptote at $x = 3$. 2. Horizontal asymptotes describe the behavior of $f(x)$ as $x$ approaches $\pm \infty$. Specifically, $\lim_{x \to -\infty} f(x)$ and $\lim_{x \to \infty} f(x)$ must be either $-4$ or $6$. 3. A vertical asymptote at $x=3$ means $\lim_{x \to 3} f(x)$ does not exist and tends to $\pm \infty$. 4. Now, analyze each statement: - Statement 1: $\lim_{x \to -\infty} f(x) = -4$, $\lim_{x \to \infty} f(x) = 6$, and $\lim_{x \to 3} f(x) = 0$. - The limits at $\pm \infty$ match the horizontal asymptotes. - But $\lim_{x \to 3} f(x) = 0$ contradicts the vertical asymptote at $x=3$ because the limit must not exist or be infinite. - Statement 2: $\lim_{x \to -\infty} f(x) = 6$, $\lim_{x \to \infty} f(x) = -4$, and $\lim_{x \to 3} f(x)$ does not exist. - The limits at $\pm \infty$ match the horizontal asymptotes (order can be reversed). - The limit at $x=3$ does not exist, consistent with the vertical asymptote. - Statement 3: $\lim_{x \to \infty} f(x) = 3$, $\lim_{x \to -4} f(x) = -\infty$, and $\lim_{x \to 6} f(x) = \infty$. - $\lim_{x \to \infty} f(x) = 3$ contradicts the horizontal asymptotes. - Limits as $x \to -4$ or $6$ are not relevant since these are $y$-values of asymptotes, not $x$-values. - Statement 4: $\lim_{x \to \infty} f(x) = 3$, $\lim_{x \to -4} f(x) = \infty$, and $\lim_{x \to 6} f(x) = -\infty$. - Same issues as Statement 3. 5. Therefore, only Statement 2 could be true. Final answer: Statement 2 is consistent with the given asymptotes.