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 $1-\frac{1}{2}=\frac{1}{2}$
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 $$Q=\lim_{z\to\infty}{f^{-1}(z)\over z^{1/3}}=\lim_{x\to\infty}{f^{-1}\bigl(f(x)\bigr)\over \bigl(f(x)\bigr)^{1/3}}=\lim_{x\to\infty}{x\over 2x\bigl(1+{3\over 8x^2}\bigr)^{1/3}}={1\over2}\ .$$
let $$f^{-1}(8x) = z, f^{-1}(x) = y.$$ then $$8x = f(z) = 8z^3+3z \to z = 2x^{1/3}+\cdots\tag 1$$ in the same way $$y = x^{1/3} + \cdots \tag 2$$ therefore $$ \lim_{x \to \infty}\dfrac {f^{-1}(8x)-f^{-1}(x)}{x^{1/3}} =\lim_{x \to \infty}\frac{z-y}{x^{1/3}} = 1.$$
Well. Both solutions of abel and Chilango are the same and both contains different misprint. I will give correction for both.
For Chilango. The number $1$ inside the first parenthesis should be $8$ $$z^{3}\left ( \color{red}{1}+\frac{3}{z^{2}} \right )=8x\Rightarrow z\left ( 1+\frac{3}{z^{2}} \right )^{1/3}=2x^{1/3}$$ Correcting this one obtains the final limit equal to $\frac{1}{2}$.
For abel. His misprint is the coefficient of $z^3$ it should be $8$ but not $1$ $$8x = f(z) = \color{red}{8}z^3+3z \to z = \color{red}{1}x^{1/3}+\cdots\tag 1$$ Taking into account this correction the limit is the same : $\frac{1}{2}$.
A simple approach:
Asymptotically, we can write $f(x) \sim 8x^3 \implies f^{-1}(x)\sim\left(\frac{x}{8}\right)^{1/3}$. then $$\lim_{x\to \infty} \frac{f^{-1}(8x)-f^{-1}(x)}{x^{1/3}}=\lim_{x\to \infty} \frac{x^{1/3}-(x/8)^{1/3}}{x^{1/3}} =\frac{1}{2}.$$
