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 $\pm \infty$. This means: - $\lim_{x \to -\infty} f(x) = -4$ or $6$ - $\lim_{x \to \infty} f(x) = -4$ or $6$ 3. Vertical asymptotes occur where the function grows without bound as $x$ approaches a certain value. Here, at $x=3$, we have: - $\lim_{x \to 3^-} f(x) = \pm \infty$ - $\lim_{x \to 3^+} f(x) = \pm \infty$ 4. Since there are two horizontal asymptotes, one must be the limit as $x \to -\infty$ and the other as $x \to \infty$. So possible limits are: $$\lim_{x \to -\infty} f(x) = -4, \quad \lim_{x \to \infty} f(x) = 6$$ or $$\lim_{x \to -\infty} f(x) = 6, \quad \lim_{x \to \infty} f(x) = -4$$ 5. At the vertical asymptote $x=3$, the limits from the left and right can be $+\infty$ or $-\infty$ independently, so: $$\lim_{x \to 3^-} f(x) = \infty \text{ or } -\infty, \quad \lim_{x \to 3^+} f(x) = \infty \text{ or } -\infty$$ 6. Therefore, statements about limits that could be true include: - $\lim_{x \to -\infty} f(x) = -4$ and $\lim_{x \to \infty} f(x) = 6$ - $\lim_{x \to -\infty} f(x) = 6$ and $\lim_{x \to \infty} f(x) = -4$ - $\lim_{x \to 3^-} f(x) = \infty$ or $-\infty$ - $\lim_{x \to 3^+} f(x) = \infty$ or $-\infty$ These are consistent with the given asymptotes.