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This relates to a previous MSE question involving an asymptotic limit:

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

Here, we seek the near-zero limit of the same expression.

So the new question is $f:R\to R, f(x)=8x^3+3x$, then find $$\lim_{x\to 0} \frac{f^{-1}(8x)-f^{-1}(x)}{x}.$$ Near zero $3x$ term ,so we take $f(x)\sim 3x \implies f^{-1}(x)\sim\frac{x}{3}. $ Hence, $$\lim_{x\to 0} \frac{f^{-1}(8x)-f^{-1}(x)}{x}=\lim_{x\to 0} \frac{8x/3-x/3}{x}=\frac{7}{3}.$$

The question here is question is whether the problem and solution are correct and what could be alternate ways of solving it?

EDIT: More rigorously as per the suggestion of @Paramanand Singh, notet that if $y=f(x)$ and inverse exists then $$\frac{d f^{-1}(y)}{dy}|_{y=y_0}=\frac{1}{f(x_0)}, y_0=f(x_0)$$ This limit can be seen as due to L-Hospital, wherein we can use $(f^{-1})'(8x)=\frac{8}{24x^2+3}=\frac{1}{3}$, when ${x \to 0}$

Z Ahmed
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    Your limit is essentially a derivative $g'(0)$ where $g(x) =f^{-1}(8x)-f^{-1}(x)$. Use the rules of derivatives combined with $f'(0)=3$ – Paramanand Singh Jan 15 '23 at 01:15
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    Your own approach needs to be made rigorous by writing $f'(0)=3\implies (f^{-1})'(0)=1/3$ so that $f^{-1}(x)=(x/3)+o(x)$ and then the limit is $7/3$. – Paramanand Singh Jan 15 '23 at 01:21

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