A function is said to have a horizontal asymptote if either the limit at infinity exists or the limit at negative infinity exists. Show that each of the following functions has a horizontal asymptote by calculating the given limit. \( \lim _{x \rightarrow \infty} \frac{-7 x}{11+2 x}=\square \) \( \lim _{x \rightarrow-\infty} \frac{4 x-6}{x^{3}+3 x-2}=\square \) \( \lim _{x \rightarrow \infty} \frac{x^{2}-7 x-10}{13-12 x^{2}}=\square \)
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To find the horizontal asymptotes of the given functions, we will compute the limits as specified. 1. For \( \lim _{x \rightarrow \infty} \frac{-7 x}{11 + 2 x} \): \[ \lim_{x \to \infty} \frac{-7x}{11 + 2x} = \lim_{x \to \infty} \frac{-7}{\frac{11}{x} + 2} = \frac{-7}{0 + 2} = -\frac{7}{2}. \] 2. For \( \lim _{x \rightarrow -\infty} \frac{4 x-6}{x^{3}+3 x-2} \): \[ \lim_{x \to -\infty} \frac{4x - 6}{x^3 + 3x - 2} = \lim_{x \to -\infty} \frac{4 - \frac{6}{x}}{x^2 + \frac{3}{x} - \frac{2}{x^2}} = \frac{4}{-\infty} = 0. \] 3. For \( \lim _{x \rightarrow \infty} \frac{x^{2}-7 x-10}{13-12 x^{2}} \): \[ \lim_{x \to \infty} \frac{x^2 - 7x - 10}{13 - 12x^2} = \lim_{x \to \infty} \frac{1 - \frac{7}{x} - \frac{10}{x^2}}{\frac{13}{x^2} - 12} = \frac{1 - 0 - 0}{0 - 12} = -\frac{1}{12}. \] Putting it all together, we have: \[ \lim _{x \rightarrow \infty} \frac{-7 x}{11 + 2 x} = -\frac{7}{2}, \quad \lim _{x \rightarrow -\infty} \frac{4 x - 6}{x^{3} + 3 x - 2} = 0, \quad \lim _{x \rightarrow \infty} \frac{x^{2} - 7 x - 10}{13 - 12 x^{2}} = -\frac{1}{12}. \]
