Question

evaluate \\(\lim_{x \to \infty} f(x)\\) and \\(\lim_{x \to -\infty} f(x)\\) for the following function. then give the horizontal asymptote(s) of \\(f\\) (if any). \\(f(x) = \frac{2x^3 + 4}{2x^3 + \sqrt{16x^6 + 3}}\\) ... evaluate \\(\lim_{x \to -\infty} f(x)\\) select the correct choice and, if necessary, fill in the answer box to complete your choice. a. \\(\lim_{x \to -\infty} f(x) = -1\\) (type an integer or a simplified fraction.) b. the limit does not exist and is neither \\(\infty\\) nor \\(-\infty\\). ... identify the horizontal asymptote(s) select the correct choice and, if necessary, fill in the answer box(es) to complete your choice. a. the function has two horizontal asymptotes. the top asymptote is \\(\square\\) and the bottom asymptote is \\(\square\\) (type equations. use integers or fractions for any numbers in the equations.) b. the function has one horizontal asymptote. \\(\square\\) (type an equation. use integers or fractions for any numbers in the equation.) c. the function has no horizontal asymptotes.

Explanation:

Step1: Evaluate $\lim_{x\to\infty} f(x)$

Divide numerator and denominator by $x^3$ (since $x^3$ is the highest power in numerator, and $\sqrt{16x^6}=4x^3$ for $x>0$):
$$f(x)=\frac{2x^3 + 4}{2x^3+\sqrt{16x^6 + 3}}=\frac{\frac{2x^3}{x^3}+\frac{4}{x^3}}{\frac{2x^3}{x^3}+\frac{\sqrt{16x^6 + 3}}{x^3}}=\frac{2+\frac{4}{x^3}}{2+\sqrt{\frac{16x^6 + 3}{x^6}}}=\frac{2+\frac{4}{x^3}}{2+\sqrt{16+\frac{3}{x^6}}}$$
As $x\to\infty$, $\frac{4}{x^3}\to0$ and $\frac{3}{x^6}\to0$:
$$\lim_{x\to\infty} f(x)=\frac{2 + 0}{2+\sqrt{16+0}}=\frac{2}{2 + 4}=\frac{2}{6}=\frac{1}{3}$$

Step2: Evaluate $\lim_{x\to-\infty} f(x)$

For $x<0$, $x^3<0$, so $\sqrt{16x^6}=|x^3|\sqrt{16}= -x^3\cdot4$ (since $x^3$ is negative, $|x^3|=-x^3$). Divide numerator and denominator by $x^3$ (note $x^3$ is negative, so dividing by $x^3$ flips signs in square root):
$$f(x)=\frac{2x^3 + 4}{2x^3+\sqrt{16x^6 + 3}}=\frac{\frac{2x^3}{x^3}+\frac{4}{x^3}}{\frac{2x^3}{x^3}+\frac{\sqrt{16x^6 + 3}}{x^3}}=\frac{2+\frac{4}{x^3}}{2+\sqrt{\frac{16x^6 + 3}{x^6}}}=\frac{2+\frac{4}{x^3}}{2-\sqrt{16+\frac{3}{x^6}}}$$
As $x\to-\infty$, $\frac{4}{x^3}\to0$ and $\frac{3}{x^6}\to0$:
$$\lim_{x\to-\infty} f(x)=\frac{2 + 0}{2-\sqrt{16+0}}=\frac{2}{2 - 4}=\frac{2}{-2}=-1$$

Step3: Determine Horizontal Asymptotes

Horizontal asymptotes occur where $\lim_{x\to\pm\infty} f(x)$ exist. We have $\lim_{x\to\infty} f(x)=\frac{1}{3}$ and $\lim_{x\to-\infty} f(x)=-1$. So there are two horizontal asymptotes: $y = \frac{1}{3}$ (from $x\to\infty$) and $y=-1$ (from $x\to-\infty$).

Answer:

• $\lim_{x\to\infty} f(x)=\boldsymbol{\frac{1}{3}}$
• $\lim_{x\to-\infty} f(x)=\boldsymbol{-1}$
• Horizontal asymptotes: $\boldsymbol{y = \frac{1}{3}}$ and $\boldsymbol{y = -1}$ (so option A is correct for horizontal asymptotes)