If $f(x) = 8x^3+3x$ then, $$\lim_{x \to \infty} \frac{f^{-1}(8x)-f^{-1}(x)}{x^{1/3}}$$ is?

My attempt: It is clear that the function cannot be easily inverted. So, there must be something in the limit given itself that may simplify the problem.

Honestly, I have no clue what to do here.

There are a few things which I could see is that the function has only 1 root(i.e 0) and is bijective on $x \in R$. But that gave no benefit except showing that the inverse of the function exists.

Any help would be appreciated.

  • $\begingroup$ The searched Limit is given by $$\frac{1}{2}$$ $\endgroup$ – Dr. Sonnhard Graubner Jan 6 at 15:29
  • $\begingroup$ Can you please explain how did you get this result? $\endgroup$ – Tony Jan 6 at 18:12
  • $\begingroup$ Nice question. +1 $\endgroup$ – Paramanand Singh Jan 7 at 8:33

Let $f(a) =x, f(b) =8x$ so that $a, b$ are functions of $x$ which tend to infinity with $x$ and $$8a^3+3a=x,8b^3+3b=8x\tag{1}$$ On subtraction we get $$(b-a)(8a^2+8b^2+8ab+3)=7x$$ and we are supposed to evaluate the limit of $(b-a) /x^{1/3}$ which is same as that of $$\frac{7x^{2/3}}{8(a^2+b^2+ab)}\tag{2}$$ At the same time note that $(1)$ implies $$\frac{b(8b^2+3)}{a(8a^2+3)}=8$$ and taking limits as $x\to\infty$ of the above equation we note that $a, b$ also tend to $\infty$ and hence we have $$\lim_{x\to\infty} \frac{b^3}{a^3}=8$$ or $b/a\to 2$. Further note that from $(1)$ we have $x/b^3\to 1$. Using these limits we can immediately see that the limit of $(2)$ is $1/2$.

  • $\begingroup$ what happened with 3 in the denominator of (2)? $\endgroup$ – user376343 Jan 7 at 12:14
  • $\begingroup$ @user376343: well $(8a^2+8b^2+8ab+3)/(8a^2+8b^2+8ab)\to 1$ as $a, b$ tend to $\infty $. You can see similar logic applied to get limit of $b^3/a^3$ later in the answer. You can see that there is no need to find inverse function and basic algebra and limit laws are needed. $\endgroup$ – Paramanand Singh Jan 7 at 13:09

Your limit is equivalent to $$ \lim_{b \to \infty} \frac{f^{-1}(8f(b))-f^{-1}(f(b))}{f(b)^{1/3}}=\lim_{b \to \infty} \frac{f^{-1}(8f(b))-b}{(8b^3+3b)^{1/3}} $$ [see also this answer for a similar problem].

Now, define $g(b)$ as $$ 8f(b)=f(g(b))\qquad (\star)\ , $$ then your limit is $$ \lim_{b \to \infty} \frac{g(b)-b}{(8b^3+3b)^{1/3}} $$ and all you need is the linear behaviour $g(b)\sim \theta b$ for $b\to\infty$, to conclude that your limit is $(\theta-1)/2$.

From $(\star)$ $$ 8(8b^3+3b)=8g^3+3g\ , $$ and replacing with the ansatz $g(b)\sim \theta b$, we see that this ansatz is only compatible with the above if $\theta=2$.


Consider the equation $$y=8x^3+3x$$ As you say, there is only one real root which is given by $$x(y)=\frac{1}{2} \left(\frac{\sqrt[3]{\sqrt{2} \sqrt{2 y^2+1}+2 y}}{2^{2/3}}-\frac{1}{\sqrt[3]{2} \sqrt[3]{\sqrt{2} \sqrt{2 y^2+1}+2 y}}\right)$$ Now, tedious but doable,for each piece, use Taylor expansion for infinitely large values of $y$ to get $$x(8y)=\sqrt[3]{y}-\frac{\sqrt[3]{\frac{1}{y}}}{8}+O\left(\frac{1}{y^{4/3}}\right) $$ $$x(y)=\frac{\sqrt[3]{y}}{2}-\frac{\sqrt[3]{\frac{1}{y}}}{4}+O\left(\frac{1}{y^{4/3}}\right)$$ $$x(8y)-x(y)=\frac{\sqrt[3]{y}}{2}+\frac{\sqrt[3]{\frac{1}{y}}}{8}+O\left(\frac{1}{y^{4/3}}\right)$$ will show the limit and also how it is approached.


May be tricky but without Taylor series for the root of the cubic equation. Let us consider that, for large values of $x$ $$8x^3+3x=8 x^3+3 x+O\left(\frac{1}{x}\right)$$ Now, series reversion gives $$x(y)=\frac{\sqrt[3]{y}}{2}-\frac{\sqrt[3]{\frac{1}{y}}}{4}+O\left(\frac{1}{y}\right)$$ and continue for the same result.

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    $\begingroup$ But this was a question given to me in an exam which only gives exactly 120 sec to solve each question. Moreover, Taylor series expansion is not even in my course. So, definitely, there might be some other way for solving this problem. Which I am unable to find. $\endgroup$ – Tony Jan 6 at 18:12

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