Answer
$\lim\limits_{x \to \infty} f(x)=\dfrac{1}{4+\sqrt 3}$
$\lim\limits_{x \to -\infty} f(x)=\dfrac{1}{4+\sqrt 3}$
Work Step by Step
We are given the function:
$f(x)=\dfrac{\sqrt[3]{x^6+8}}{4x^2+\sqrt{3x^4+1}}$
Determine $\lim\limits_{x \to \infty} f(x)$:
$\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{\sqrt[3]{x^6+8}}{4x^2+\sqrt{3x^4+1}}$
$=\lim\limits_{x \to \infty} \dfrac{\dfrac{\sqrt[3]{x^6+8}}{x^2}}{\dfrac{4x^2+\sqrt{3x^4+1}}{x^2}}$
$=\lim\limits_{x \to \infty} \dfrac{\sqrt[3]{\dfrac{x^6+8}{x^6}}}{\dfrac{4x^2}{x^2}+\sqrt{\dfrac{3x^4+1}{x^4}}}$
$=\lim\limits_{x \to \infty} \dfrac{\sqrt[3]{\dfrac{x^6}{x^6}+\dfrac{8}{x^6}}}{4+\sqrt{\dfrac{3x^4}{x^4}+\dfrac{1}{x^4}}}$
$=\lim\limits_{x \to \infty} \dfrac{\sqrt[3]{1+\dfrac{8}{x^6}}}{4+\sqrt{3+\dfrac{1}{x^4}}}$
$=\dfrac{1}{4+\sqrt 3}$
Determine $\lim\limits_{x \to -\infty} f(x)$:
$\lim\limits_{x \to -\infty} f(x)=\lim\limits_{x \to -\infty} \dfrac{\sqrt[3]{x^6+8}}{4x^2+\sqrt{3x^4+1}}$
$=\lim\limits_{x \to -\infty} \dfrac{\dfrac{\sqrt[3]{x^6+8}}{x^2}}{\dfrac{4x^2+\sqrt{3x^4+1}}{x^2}}$
$=\lim\limits_{x \to -\infty} \dfrac{\sqrt[3]{\dfrac{x^6+8}{x^6}}}{\dfrac{4x^2}{x^2}+\sqrt{\dfrac{3x^4+1}{x^4}}}$
$=\lim\limits_{x \to -\infty} \dfrac{\sqrt[3]{\dfrac{x^6}{x^6}+\dfrac{8}{x^6}}}{4+\sqrt{\dfrac{3x^4}{x^4}+\dfrac{1}{x^4}}}$
$=\lim\limits_{x \to -\infty} \dfrac{\sqrt[3]{1+\dfrac{8}{x^6}}}{4+\sqrt{3+\dfrac{1}{x^4}}}$
$=\dfrac{1}{4+\sqrt 3}$
