If $f(x)=8x^3+3x$ , $x\in\mathbb{R}$, how do I find $\lim_{x \to \infty}\frac {f^{-1}(8x)-f^{-1}(x)}{x^{1/3}}$? 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 
$$\lim_{x \to \infty}\dfrac {f^{-1}(8x)-f^{-1}(x)}{x^{1/3}}?$$

I got something different than the one in the prevous answer:
$f^{-1}(8x)=z$ and $f^{-1}(x)=y$ so
$8x=8z^{3}+3z$ and $x=8y^{3}+3y$
Therefore,
$8z^{3}\left ( 1+\frac{3}{z^{2}} \right )=8x\Rightarrow z\left ( 1+\frac{3}{z^{2}} \right )^{1/3}=x^{1/3}$ and similarly
$2y\left ( 1+\frac{3}{8y^{2}} \right )^{1/3}=x^{1/3}$. 
Now, substituting, then factoring and cancelling $x^{1/3}$, we get
$\dfrac{z-y}{x^{1/3}}=\dfrac{\dfrac{x^{1/3}}{\left ( 1+\dfrac{3}{z^{2}} \right )^{1/3}}-\left (\dfrac{ x^{1/3}}{2\left ( 1+\dfrac{3}{8y^{2}} \right )^{1/3}}\right )}{x^{1/3}}=\\\dfrac{1}{\left ( 1+\dfrac{3}{z^{2}} \right )^{1/3}}-\left (\dfrac{1}{2\left ( 1+\dfrac{3}{8y^{2}} \right )^{1/3}}\right )$. 
As $x\rightarrow \infty $, $z\rightarrow \infty$ so upon taking limits, we obtain $1-\frac{1}{2}=\frac{1}{2}$
