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.\
\\( \bigcirc \\) a. \\( \lim_{x \to \infty} f(x) = \square \\) (type an integer or a simplified fraction.)\
\\( \bigcirc \\) b. the limit does not exist and is neither \\( \infty \\) nor \\( -\infty \\).

Explanation:

Step1: Divide numerator and denominator by \(x^3\)

For \(x \to \infty\), \(x^3>0\), so \(\sqrt{16x^6 + 3}=x^3\sqrt{16+\frac{3}{x^6}}\). Divide numerator \(2x^3 + 4\) and denominator \(2x^3+\sqrt{16x^6 + 3}\) by \(x^3\):
\[
f(x)=\frac{\frac{2x^3}{x^3}+\frac{4}{x^3}}{\frac{2x^3}{x^3}+\frac{\sqrt{16x^6 + 3}}{x^3}}=\frac{2+\frac{4}{x^3}}{2+\sqrt{16+\frac{3}{x^6}}}
\]

Step2: Take the limit as \(x \to \infty\)

As \(x \to \infty\), \(\frac{4}{x^3}\to 0\) and \(\frac{3}{x^6}\to 0\). So:
\[
\lim_{x \to \infty}f(x)=\frac{2 + 0}{2+\sqrt{16+0}}=\frac{2}{2 + 4}=\frac{2}{6}=\frac{1}{3}
\]

Answer:

A. \(\lim\limits_{x\to\infty} f(x)=\frac{1}{3}\)