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. $$ \lim _{x \rightarrow \infty} \frac{-6 x}{10+2 x}= $$ $$ \lim _{x \rightarrow-\infty} \frac{15 x-12}{x^{3}+14 x-8}= $$ $$ \lim _{x \rightarrow \infty} \frac{x^{2}-6 x-6}{2-8 x^{2}}= $$ $$ \lim _{x \rightarrow \infty} \frac{\sqrt{x^{2}+3 x}}{4-2 x}= $$

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We begin by analyzing each limit separately.

1. For 
$$
\lim_{x\to\infty} \frac{-6x}{10+2x},
$$
divide the numerator and denominator by $$ x $$:
$$
\frac{-6x}{10+2x} = \frac{-6}{\frac{10}{x}+2}.
$$
As $$ x\to\infty $$, $$\frac{10}{x}\to 0$$, so the limit becomes
$$
\frac{-6}{0+2} = -3.
$$

2. For 
$$
\lim_{x\to-\infty} \frac{15x-12}{x^3+14x-8},
$$
notice that the highest power in the numerator is $$ x $$ and in the denominator is $$ x^3 $$. Dividing numerator and denominator by $$ x^3 $$ gives
$$
\frac{15x-12}{x^3+14x-8} = \frac{\frac{15}{x^2}-\frac{12}{x^3}}{1+\frac{14}{x^2}-\frac{8}{x^3}}.
$$
As $$ x\to-\infty $$, the terms $$\frac{15}{x^2}$$, $$\frac{12}{x^3}$$, $$\frac{14}{x^2}$$, and $$\frac{8}{x^3}$$ all tend to 0, so we obtain
$$
\frac{0-0}{1+0-0} = 0.
$$

3. For 
$$
\lim_{x\to\infty} \frac{x^2-6x-6}{2-8x^2},
$$
the highest powers in both numerator and denominator are $$x^2$$. Divide numerator and denominator by $$x^2$$:
$$
\frac{x^2-6x-6}{2-8x^2} = \frac{1-\frac{6}{x}-\frac{6}{x^2}}{\frac{2}{x^2}-8}.
$$
Taking $$ x\to\infty $$, the fractions with $$ x $$ in the denominator vanish:
$$
\frac{1-0-0}{0-8} = -\frac{1}{8}.
$$

4. For 
$$
\lim_{x\to\infty} \frac{\sqrt{x^2+3x}}{4-2x},
$$
first simplify the numerator. Factor $$ x^2 $$ inside the square root:
$$
\sqrt{x^2+3x} = \sqrt{x^2\left(1+\frac{3}{x}\right)} = x\sqrt{1+\frac{3}{x}},
$$
since for $$ x\to\infty $$ we have $$ x>0 $$. Thus, the expression becomes
$$
\frac{x\sqrt{1+\frac{3}{x}}}{4-2x}.
$$
Factor $$-2x$$ out of the denominator:
$$
4-2x = -2x\left(1-\frac{2}{x}\right).
$$
Then,
$$
\frac{x\sqrt{1+\frac{3}{x}}}{-2x\left(1-\frac{2}{x}\right)} = \frac{\sqrt{1+\frac{3}{x}}}{-2\left(1-\frac{2}{x}\right)}.
$$
Taking the limit as $$ x\to\infty $$ where $$\sqrt{1+\frac{3}{x}}\to 1$$ and $$1-\frac{2}{x}\to 1$$, we get
$$
\frac{1}{-2} = -\frac{1}{2}.
$$

Thus, the horizontal asymptotes are determined by:

$$
\lim_{x\to\infty} \frac{-6x}{10+2x}=-3,
$$
$$
\lim_{x\to-\infty} \frac{15x-12}{x^3+14x-8}=0,
$$
$$
\lim_{x\to\infty} \frac{x^2-6x-6}{2-8x^2}=-\frac{1}{8},
$$
$$
\lim_{x\to\infty} \frac{\sqrt{x^2+3x}}{4-2x}=-\frac{1}{2}.
$$
The horizontal asymptotes are: 1. $$ y = -3 $$ 2. $$ y = 0 $$ 3. $$ y = -\frac{1}{8} $$ 4. $$ y = -\frac{1}{2} $$

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.

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