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Is my demonstration correct?

Let $f, g$ infinitely differentiable increasing functions. Suposse that:

$$\star \ \ \ \ \ f(0) = g(0) = 0 $$

$$\star \ \ \ \ \ f'(0) = g'(0) = 1 $$

$$\star \ \ \ \ \ \lim\limits_{x \rightarrow 0} \dfrac{f(x)}{x} = 1 $$

$$\star \ \ \ \ \ f(-x) = -f(x), g(-x) = -g(x) $$

$$\star \ \ \ \ \ f(x) \neq g(x) \ \ \ \forall x \neq 0 $$

So $$\lim\limits_{x \rightarrow 0}\dfrac{f(x) - g(x)}{g^{-1}(x) - f^{-1}(x)} = 1 $$

My possible demonstration: We can approximate f(x) next to $x_0$ by

$$(1) \ \ \ f(x) \approx f'(x_0)x + (f(x_0) - x_0f'(x_0)) = f'(x_0)x + b(x_0) $$

where $b(x_0) = (f(x_0) - x_0f'(x))$. Replacing x by $f^{-1}(x)$ in (1) we got:

$$(2) \ \ \ f^{-1}(x) = \dfrac{x - b(x_0)}{f'(x_0)} $$

(same about $g$, $g(x) = g'(x_0)x + (g(x_0) - x_0g'(x_0)) = g'(x_0)x + c(x_0))$,

so

$$\dfrac{f(x) - g(x)}{g^{-1}(x) - f^{-1}(x)} \approx \dfrac{f'(x_0)x + b(x_0) - (g'(x_0)x + c(x_0))}{\dfrac{x - c(x_0)}{g'(x_0)}-\dfrac{x - b(x_0)}{f'(x_0)}} $$

But $$\lim\limits_{x_0 \rightarrow 0} f'(x_0) = \lim\limits_{x_0 \rightarrow 0} g'(x_0) =1$$

So we got

$$\dfrac{f'(x_0)x + b(x_0) - (g'(x_0)x + c(x_0))}{\dfrac{x - c(x_0)}{g'(x_0)}-\dfrac{x - b(x_0)}{f'(x_0)}} \approx \dfrac{b(x_0) - c(x_0)}{-c(x_0) + b(x_0)} = 1$$

It's wrong? How can I be more accurate?

I want to prove for $f(x) = \sin(\tan(x)), g(x) = \tan(\sin(x))$, however, I believe it is possible to generalize as above

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    You're going to be more accurate by replacing $\approx$ with something more precise, such as an $\epsilon-\delta$ notation or a Bachmann-Landau $o$ notation – FShrike Dec 27 '21 at 15:50
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    your proof is not correct! $f(x)=x$ and $g(x) = x + sgn(x)\cdot e^{-\frac1{x^2}}$ should cause problems to your proof with expansions, since they have the exact same Taylor expansion! I am not sure whether or not this is a counterexample to the initial statement though – LucaMac Dec 27 '21 at 15:51
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    You don't require $f,g$ to be infinitely differentiable; continuity of the first derivative would do. If you are writing this up as a formal proof, you'd probably want to add that you used $f'(0)=g'(0)=1\neq0$ and invoke the Inverse Function Theorem to say that $f^{-1},g^{-1}$ exist and are $C^1$ in some neighbourhood of $0$ – FShrike Dec 27 '21 at 15:52
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    Note that the second condition implies the third one by definition of derivatives, and the fourth one implies the first one. – Taladris Dec 27 '21 at 16:19
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    https://math.stackexchange.com/questions/2513708/how-many-times-do-you-have-to-use-lh%c3%b4pitals-rule/2513821#2513821 – Maxim Gilula Dec 27 '21 at 16:23
  • Thank you all, you helped me a lot. –  Dec 27 '21 at 17:46

0 Answers0