Evaluating $\lim_{x\to \infty} \frac{f^{-1}(2021x)-f^{-1}(x)}{\sqrt[2021] x}$, where $f(x)=2021x^{2021}+x+1$ 
Let $f(x)=2021x^{2021}+x+1$, and compute the following limit:
$$\lim_{x\to \infty} \frac{f^{-1}(2021x)-f^{-1}(x)}{\sqrt[2021] x}$$

My attempt: i want to use mean value theorem to $f^{-1}(x)$ then we have:
$ \frac{2020x}{((2021)^2 (η_y)^{2021}+1)(\sqrt[2021] x)}$.
And $f(η_y)=η_x$, $η_x∈(x,2021x) $. We know $η_x$ tend to 0 as x tend to $\infty$ so from the expression of $f$ we know $η_y$ is also.but we  don't know  the quotient of $η_y$ and $x$, so i think this way is fail. But i can't do it in other way.
 A: $$L=\lim_{x\to \infty} \frac{f^{-1}(2021x)}{\sqrt[2021]x}=\sqrt[2021]{2021}\lim_{u\to \infty} \frac{f^{-1}u}{\sqrt[2021]{u}}$$ now let $u=f(x)$ $$L=\sqrt[2021]{2021}\lim_{x\to \infty} \frac{x}{\sqrt[2021]{2021x^{2021}+x+1}}=1$$ now $$\tau=\lim_{x\to \infty} \frac{f^{-1}x}{\sqrt[2021]{x}}=\lim_{t\to \infty}\frac{f^{-1}(f(t))}{\sqrt[2021]{f(t)}}=\lim_{t\to \infty} \frac{t}{\sqrt[2021]{2021t^{2021}+t+1}}=\frac{1}{\sqrt[2021]{2021}}$$ therefore required value is $$L-\tau=?$$
A: Let $y=x^{1/2021}$, then $x\to\infty$ iff $y\to\infty$, so we consider the limit
$$\lim_{y\to\infty}\frac{f^{-1}(2021y^{2021})-f^{-1}(y^{2021})}{y}.$$
Note that $f$ is strictly increasing, so $f^{-1}(2021y^{2021})<y$. Note that $2021y^{2021}-f(y-1)$ is a polynomial of even degree $2020$ whose leading term has a positive coefficient $2021^2>0$, so when $|y|$ is sufficiently large, we have $f(y-1)<2021y^{2021}$ and thus $y-1<f^{-1}(2021y^{2021})$. Hence
$$0<\left|\frac{f^{-1}(2021y^{2021})-y}{y}\right|<\frac1{|y|}, \qquad |y|>>1.$$
Therefore
$$\lim_{y\to\infty}\frac{f^{-1}(2021y^{2021})}{y}=1.$$
And so
$$\lim_{y\to\infty}\frac{f^{-1}(y^{2021})}{y}=\lim_{y\to\infty}\frac{f^{-1}(2021(2021^{-1/2021}y^{2021}))}{y}=2021^{-1/2021}.$$
Finally,
$$\lim_{y\to\infty}\frac{f^{-1}(2021y^{2021})-f^{-1}(y^{2021})}{y}=1-2021^{-1/2021}=1-\left(\frac1{2021}\right)^{1/2021}.$$
A: For an arbitrary $\epsilon>0$ we have
$$
2021x^{2021}<f(x)<(2021+\epsilon)x^{2021}
$$
which holds for sufficiently large $x$. Hence
$$
(\frac{x}{2021+\epsilon})^\frac{1}{2021}<f^{-1}(x)<(\frac{x}{2021})^\frac{1}{2021},
$$
from which
$$
(\frac{x}{1+\epsilon_1})^\frac{1}{2021}<f^{-1}(2021x)<(x)^\frac{1}{2021},
$$
where $\epsilon_1=\epsilon/2021$ and we can write
$$
(\frac{2021}{2021+\epsilon})^\frac{1}{2021}-(\frac{1}{2021})^\frac{1}{2021}
<
\frac{f^{-1}(2021x)-f^{-1}(x)}{x^{\frac{1}{2021}}}<1-(\frac{1}{2021+\epsilon})^\frac{1}{2021}.
$$
Since the above bounds hold for any $\epsilon>0$, the limit must be
$
1-(\frac{1}{2021})^\frac{1}{2021}\approx 0.00375904653
$
.
