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) = \frac{1}{3} \\) (type an integer or a simplified fraction.)
b. the limit does not exist and is neither \\( \infty \\) nor \\( -\infty \\).

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) = \square \\) (type an integer or a simplified fraction.)
b. the limit does not exist and is neither \\( \infty \\) nor \\( -\infty \\).

Explanation:

Step1: Analyze \( x \to -\infty \), simplify \( \sqrt{16x^6 + 3} \)

For \( x \to -\infty \), \( x^3 \) is negative. \( \sqrt{16x^6 + 3} = |4x^3| = -4x^3 \) (since \( x^3 < 0 \) when \( x \to -\infty \)).

Step2: Substitute into \( f(x) \)

\( f(x)=\frac{2x^3 + 4}{2x^3 + \sqrt{16x^6 + 3}}=\frac{2x^3 + 4}{2x^3 - 4x^3}=\frac{2x^3 + 4}{-2x^3} \)

Step3: Divide numerator and denominator by \( x^3 \)

\( \lim_{x \to -\infty} \frac{2x^3 + 4}{-2x^3}=\lim_{x \to -\infty} \frac{2 + \frac{4}{x^3}}{-2} \)

Step4: Evaluate the limit

As \( x \to -\infty \), \( \frac{4}{x^3} \to 0 \). So \( \lim_{x \to -\infty} \frac{2 + 0}{-2} = -1 \)

Answer:

For \( \lim_{x \to -\infty} f(x) \), the correct choice is A, and \( \lim_{x \to -\infty} f(x) = -1 \)