TITLE: If $f(x)=8x^3+3x$ , $x\in\mathbb{R}$, how do I find $\lim_{x \to \infty}\frac {f^{-1}(8x)-f^{-1}(x)}{x^{1/3}}$?
QUESTION [5 upvotes]: Let $f: \mathbb R \to \mathbb R$ be defined as $f(x)=8x^3+3x$. Then $f$ is continuous , strictly increasing, and 
$\lim _{x\to \infty}f(x)=\infty , \lim_{x \to -\infty}f(x)=-\infty$ , so $f$ is injective and surjective . How do I compute 
$$\lim_{x \to \infty}\dfrac {f^{-1}(8x)-f^{-1}(x)}{x^{1/3}}?$$

REPLY [3 votes]: One has
$$\lim_{y\to\infty}{f^{-1}(8y)\over y^{1/3}}=2\lim_{y\to\infty}{f^{-1}(8y)\over(8y)^{1/3}}=2\lim_{z\to\infty}{f^{-1}(z)\over z^{1/3}}\ .$$
Therefore the quantity in $Q$ in question is equal to
$$Q=\lim_{z\to\infty}{f^{-1}(z)\over z^{1/3}}=\lim_{x\to\infty}{f^{-1}\bigl(f(x)\bigr)\over \bigl(f(x)\bigr)^{1/3}}=\lim_{x\to\infty}{x\over 2x\bigl(1+{3\over 8x^2}\bigr)^{1/3}}={1\over2}\ .$$