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. $$ \lim _{x \rightarrow \infty} \frac{-6 x}{10+2 x}=-3 $$ $$ \lim _{x \rightarrow-\infty} \frac{15 x-12}{x^{3}+14 x-8}=0 $$ $$ \lim _{x \rightarrow \infty} \frac{x^{2}-6 x-6}{2-8 x^{2}}=-1 / 8 $$ $$ \lim _{x \rightarrow \infty} \frac{\sqrt{x^{2}+3 x}}{4-2 x}= $$ $$ \frac{\sqrt{x^{2}+3 x}}{4}= $$

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### 1. Limit: $$\displaystyle \lim _{x \rightarrow \infty} \frac{-6 x}{10+2 x}$$

Factor $$x$$ from the denominator:
$$
\frac{-6x}{10+2x}=\frac{-6x}{x\left( \frac{10}{x}+2 \right)}=\frac{-6}{\frac{10}{x}+2}.
$$
Taking the limit as $$x\to\infty$$, note that $$\frac{10}{x}\to 0$$. Hence,
$$
\lim_{x\to\infty}\frac{-6}{\frac{10}{x}+2}=\frac{-6}{0+2}=-3.
$$
Thus, the function has a horizontal asymptote at $$y=-3$$.

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### 2. Limit: $$\displaystyle \lim _{x \rightarrow -\infty} \frac{15x-12}{x^{3}+14x-8}$$

Observe that the degree of the polynomial in the numerator is 1, while that in the denominator is 3. For large $$|x|$$ the highest power dominates, so we have:
$$
\lim_{x\to -\infty}\frac{15x-12}{x^3+14x-8}\sim \lim_{x\to -\infty}\frac{15x}{x^3}=\lim_{x\to -\infty}\frac{15}{x^2}=0.
$$
Thus, the function has a horizontal asymptote at $$y=0$$.

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### 3. Limit: $$\displaystyle \lim _{x \rightarrow \infty} \frac{x^{2}-6 x-6}{2-8 x^{2}}$$

Divide numerator and denominator by $$x^2$$, the highest power in the expression:
$$
\frac{x^2-6x-6}{2-8x^2}=\frac{x^2\left(1-\frac{6}{x}-\frac{6}{x^2}\right)}{x^2\left(\frac{2}{x^2}-8\right)}
=\frac{1-\frac{6}{x}-\frac{6}{x^2}}{\frac{2}{x^2}-8}.
$$
Taking the limit as $$x\to\infty$$ (where $$\frac{6}{x}\to 0$$, $$\frac{6}{x^2}\to 0$$, $$\frac{2}{x^2}\to 0$$), we get:
$$
\lim_{x\to\infty}\frac{1-0-0}{0-8}=\frac{1}{-8}=-\frac{1}{8}.
$$
Thus, the function has a horizontal asymptote at $$y=-\frac{1}{8}$$.

---

### 4. Limit: $$\displaystyle \lim _{x \rightarrow \infty} \frac{\sqrt{x^{2}+3 x}}{4-2 x}$$

Start by rewriting the square root in the numerator:
$$
\sqrt{x^2+3x}=x\sqrt{1+\frac{3}{x}},\quad \text{for }x>0.
$$
Then the expression becomes:
$$
\frac{x\sqrt{1+\frac{3}{x}}}{4-2x}.
$$
Rewrite the denominator by factoring out $$-2x$$:
$$
4-2x=-2x\left(1-\frac{4}{2x}\right)=-2x\left(1-\frac{2}{x}\right).
$$
Now, the ratio simplifies to:
$$
\frac{x\sqrt{1+\frac{3}{x}}}{-2x\left(1-\frac{2}{x}\right)}
=\frac{\sqrt{1+\frac{3}{x}}}{-2\left(1-\frac{2}{x}\right)}.
$$
Taking the limit as $$x\to\infty$$ (where $$\frac{3}{x}\to 0$$ and $$\frac{2}{x}\to 0$$):
$$
\lim_{x\to\infty}\frac{\sqrt{1+\frac{3}{x}}}{-2\left(1-\frac{2}{x}\right)}
=\frac{\sqrt{1+0}}{-2(1-0)}
=\frac{1}{-2}=-\frac{1}{2}.
$$
Thus, the function has a horizontal asymptote at $$y=-\frac{1}{2}$$.

---

### 5. Expression: $$\displaystyle \frac{\sqrt{x^{2}+3 x}}{4}$$

In this case, the function is $$\frac{\sqrt{x^{2}+3x}}{4}$$. For large $$x$$ (with $$x>0$$):
$$
\sqrt{x^2+3x}=x\sqrt{1+\frac{3}{x}},
$$
so the expression becomes:
$$
\frac{x\sqrt{1+\frac{3}{x}}}{4}.
$$
As $$x\to\infty$$, the dominant behavior is:
$$
\frac{x\sqrt{1+\frac{3}{x}}}{4}\sim\frac{x}{4}\to \infty.
$$
Since the limit does not approach a finite number, the function does not have a horizontal asymptote.

---

In summary:

- $$\displaystyle \lim _{x \rightarrow \infty} \frac{-6 x}{10+2 x}=-3$$ (horizontal asymptote: $$y=-3$$)
- $$\displaystyle \lim _{x \rightarrow-\infty} \frac{15 x-12}{x^{3}+14 x-8}=0$$ (horizontal asymptote: $$y=0$$)
- $$\displaystyle \lim _{x \rightarrow \infty} \frac{x^{2}-6 x-6}{2-8 x^{2}}=-\frac{1}{8}$$ (horizontal asymptote: $$y=-\frac{1}{8}$$)
- $$\displaystyle \lim _{x \rightarrow \infty} \frac{\sqrt{x^{2}+3 x}}{4-2 x}=-\frac{1}{2}$$ (horizontal asymptote: $$y=-\frac{1}{2}$$)
- The function $$\displaystyle \frac{\sqrt{x^{2}+3 x}}{4}$$ does not have a horizontal asymptote since its limit as $$x\to\infty$$ is unbounded.
The functions have the following horizontal asymptotes: 1. $$ y = -3 $$ 2. $$ y = 0 $$ 3. $$ y = -\frac{1}{8} $$ 4. $$ y = -\frac{1}{2} $$ 5. No horizontal asymptote

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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1. Limit:

2. Limit:

3. Limit:

4. Limit:

5. Expression:

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