Question

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 \to \infty} \frac{-6x}{14 + 2x} = \\) \\(\lim_{x \to -\infty} \frac{13x - 13}{x^3 + 11x - 9} = \\) \\(\lim_{x \to \infty} \frac{x^2 - 9x - 5}{8 - 14x^2} = \\) \\(\lim_{x \to \infty} \frac{\sqrt{x^2 + 9x}}{3 - 13x} = \\) \\(\lim_{x \to -\infty} \frac{\sqrt{x^2 + 9x}}{3 - 13x} = \\) question help: \\(\square\\) message instructor

Explanation:

Step1: Solve $\boldsymbol{\lim_{x\to\infty} \frac{-6x}{14 + 2x}}$

Divide numerator and denominator by $x$:
$\lim_{x\to\infty} \frac{-6}{ \frac{14}{x} + 2 }$
As $x\to\infty$, $\frac{14}{x}\to0$. So:
$\frac{-6}{0 + 2} = -3$

Step2: Solve $\boldsymbol{\lim_{x\to-\infty} \frac{13x - 13}{x^3 + 11x - 9}}$

For large $|x|$, the highest power term dominates. Numerator is degree 1, denominator degree 3.
$\lim_{x\to-\infty} \frac{13x}{x^3} = \lim_{x\to-\infty} \frac{13}{x^2} = 0$ (since $x^2\to\infty$)

Step3: Solve $\boldsymbol{\lim_{x\to\infty} \frac{x^2 - 9x - 5}{8 - 14x^2}}$

Divide numerator and denominator by $x^2$:
$\lim_{x\to\infty} \frac{1 - \frac{9}{x} - \frac{5}{x^2}}{\frac{8}{x^2} - 14}$
As $x\to\infty$, $\frac{9}{x},\frac{5}{x^2},\frac{8}{x^2}\to0$. So:
$\frac{1 - 0 - 0}{0 - 14} = -\frac{1}{14}$

Step4: Solve $\boldsymbol{\lim_{x\to\infty} \frac{\sqrt{x^2 + 9x}}{3 - 13x}}$

Simplify $\sqrt{x^2}=|x|=x$ (since $x\to\infty$). Divide numerator and denominator by $x$:
$\lim_{x\to\infty} \frac{\sqrt{1 + \frac{9}{x}}}{\frac{3}{x} - 13}$
As $x\to\infty$, $\frac{9}{x},\frac{3}{x}\to0$. So:
$\frac{\sqrt{1 + 0}}{0 - 13} = -\frac{1}{13}$

Step5: Solve $\boldsymbol{\lim_{x\to-\infty} \frac{\sqrt{x^2 + 9x}}{3 - 13x}}$

Simplify $\sqrt{x^2}=|x|=-x$ (since $x\to-\infty$). Divide numerator and denominator by $x$ (note $x$ is negative, so $\frac{\sqrt{x^2}}{x}=\frac{-x}{x}=-1$):
$\lim_{x\to-\infty} \frac{\sqrt{1 + \frac{9}{x}}}{\frac{3}{x} - 13} \cdot \frac{-x}{x}$ (adjust for $|x|=-x$)
Wait, better: divide numerator and denominator by $-x$ (positive as $x\to-\infty$):
Numerator: $\frac{\sqrt{x^2 + 9x}}{-x} = \sqrt{\frac{x^2 + 9x}{x^2}} = \sqrt{1 + \frac{9}{x}}$ (since $-x>0$)
Denominator: $\frac{3 - 13x}{-x} = -\frac{3}{x} + 13$
As $x\to-\infty$, $\frac{9}{x},\frac{3}{x}\to0$. So:
$\frac{\sqrt{1 + 0}}{0 + 13} = \frac{1}{13}$

Answer:

• $\lim_{x\to\infty} \frac{-6x}{14 + 2x} = \boldsymbol{-3}$
• $\lim_{x\to-\infty} \frac{13x - 13}{x^3 + 11x - 9} = \boldsymbol{0}$
• $\lim_{x\to\infty} \frac{x^2 - 9x - 5}{8 - 14x^2} = \boldsymbol{-\frac{1}{14}}$
• $\lim_{x\to\infty} \frac{\sqrt{x^2 + 9x}}{3 - 13x} = \boldsymbol{-\frac{1}{13}}$
• $\lim_{x\to-\infty} \frac{\sqrt{x^2 + 9x}}{3 - 13x} = \boldsymbol{\frac{1}{13}}$