Answer
$\lim\limits_{x \to \infty} f(x)=\dfrac{2}{3}$
$\lim\limits_{x \to -\infty} f(x)=-2$
Work Step by Step
We are given the function:
$f(x)=\dfrac{4x^3+1}{2x^3+\sqrt{16x^6+1}}$
Determine $\lim\limits_{x \to \infty} f(x)$:
$\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{4x^3+1}{2x^3+\sqrt{16x^6+1}}$
$=\lim\limits_{x \to \infty}\dfrac{\dfrac{4x^3}{x^3}+\dfrac{1}{x^3}}{\dfrac{2x^3}{x^3}+\dfrac{\sqrt{16x^6+1}}{x^3}}$
$=\lim\limits_{x \to \infty}\dfrac{4+\dfrac{1}{x^3}}{2+\sqrt{\dfrac{16x^6}{x^6}+\dfrac{1}{x^6}}}$
$=\lim\limits_{x \to \infty}\dfrac{4+\dfrac{1}{x^3}}{2+\sqrt{16+\dfrac{1}{x^6}}}$
$=\dfrac{4}{2+\sqrt{16}}=\dfrac{4}{2+4}$
$=\dfrac{4}{6}=\dfrac{2}{3}$
Determine $\lim\limits_{x \to -\infty} f(x)$:
$\lim\limits_{x \to -\infty} f(x)=\lim\limits_{x \to -\infty} \dfrac{4x^3+1}{2x^3+\sqrt{16x^6+1}}$
$=\lim\limits_{x \to -\infty}\dfrac{\dfrac{4x^3}{x^3}+\dfrac{1}{x^3}}{\dfrac{2x^3}{x^3}+\dfrac{\sqrt{16x^6+1}}{x^3}}$
$=\lim\limits_{x \to -\infty}\dfrac{4+\dfrac{1}{x^3}}{2-\sqrt{\dfrac{16x^6}{x^6}+\dfrac{1}{x^6}}}$
$=\lim\limits_{x \to -\infty}\dfrac{4+\dfrac{1}{x^3}}{2-\sqrt{16+\dfrac{1}{x^6}}}$
$=\dfrac{4}{2-\sqrt{16}}=\dfrac{4}{2-4}$
$=\dfrac{4}{-2}=-2$
