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I have this limit of this form

$$f(x)^{g(x)}=e^{g(x)\ln(f(x))}$$

$$\lim _{x\to 0\color{red}{\boldsymbol -}}\left(1+x^3\right)^{1/\left((x^2+1)^4-1\right)}$$

In our case I can write in the exponent:

$${g(x)\ln(f(x))}=\frac{\ln(f(x))}{\frac1{g(x)}}$$

and I have an indeterminate form $(0/0)$ and I can apply de l'Hôpital rule. Right now I just thought to write

$$(1+x^3)=\left(1+\frac{1}{\frac1{x^3}}\right)$$ and I call $x^3=t$ but I think to obtain the exponent too long and it will be more complicated.

Théophile
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Sebastiano
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3 Answers3

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Hint:

Be simple and use equivalents near $0$: the logarithm is $$\frac{\ln(1+x^3)}{(x^2+1)^4-1}= \frac{\ln(1+x^3)}{\bigl((x^2+1)^2-1\bigr)\bigl((x^2+1)^2+1\bigr)}\sim_0\frac{x^3}{2x^2\cdot 2}=\frac x4$$ therefore…

Bernard
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    god I love infinitesimal analysis so much – tryst with freedom Aug 20 '20 at 19:56
  • Very nice, but I'm helping my cousin's son in registered in the faculty of Economics and Finance and I can't use asymptotics. +1 – Sebastiano Aug 20 '20 at 21:31
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    Are they allowed to use Taylor-Young's formula? Other idea: it's known in high school that $\lim_{u\to 0}\dfrac{\ln (1+u)}u=1$ (which is the equivalence relation without saying so). – Bernard Aug 20 '20 at 21:35
  • I just read your comment now because you didn't enter the @+ my name. The problem is that you don't have the math knowledge not having studied in a school where they don't deal with limits and integrals. Let's say the whole analysis 1. I didn't know the name of $\lim_{u\to 0}\dfrac{\ln (1+u)}u=1$ is Taylor-Young's formula. – Sebastiano Aug 20 '20 at 21:57
  • @Sebastiano (I didn't forget this time!). No,no, this limit is a high-school limit. Taylor-Young's formula expresses the remainder in Taylor's formula as $o(x^n)$, e.g. $\sin x =x-\frac{x^3}{3!}+o(x^3)$. – Bernard Aug 20 '20 at 22:06
  • @Bernard Yes of course. You're right 100%. But in Italy there are particular schools where there is not the explanation of the limits and the integrals. The schools are called technical-commercial high school. – Sebastiano Aug 20 '20 at 22:14
  • But the limit I pointed as a high-school limit? I just corresponds to the definition of the derivative of $\ln x$ at $x=1$. – Bernard Aug 20 '20 at 22:17
  • @Bernard Very kind Bernard the students of technical-commercial high school not use the limits and the derivates and the integrals. The level of the student is low. :-( – Sebastiano Aug 21 '20 at 09:03
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We can use that

$$\large{\left(1+x^3\right)^{\frac{1}{\left(x^2+1\right)^4-1}}=\left[\left(1+x^3\right)^{\frac1{x^3}}\right]^{\frac{x^3}{\left(x^2+1\right)^4-1}}}\to e^0=1$$

indeed

$$\left(x^2+1\right)^4=1+4x^2+O(x^4) \implies \frac{x^3}{\left(x^2+1\right)^4-1}= \frac{x^3}{4x^2+O(x^4)}=\frac{x}{4+O(x^2)}\to 0$$

user
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    It is true :-(..but why didn't I come up with the idea in the first place? You're absolutely right about the resolution being immediate. Thank you all very much. – Sebastiano Aug 20 '20 at 21:32
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    @Sebastiano You are welcome! I'm glad you can find that helpful. Bye – user Aug 20 '20 at 21:37
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    And I'm happy to meet very good, educated and very expert users... – Sebastiano Aug 20 '20 at 21:39
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$$A=\left(1+x^3\right)^{\frac{1}{\left(x^2+1\right)^4-1}}\implies \log(A)=\frac{1}{\left(x^2+1\right)^4-1}\log(1+x^3)$$ $$\log(A)=\frac{x^3-\frac{1}{2}x^6+\frac{1}{3}x^9+O\left(x^{12}\right)}{4 x^2+6 x^4+4 x^6+x^8}$$ Long division $$\log(A)=\frac{x}{4}-\frac{3 x^3}{8}+O\left(x^{4}\right)$$ $$A=e^{\log(A)}=1+\frac{x}{4}+\frac{x^2}{32}+O\left(x^3\right)$$

Claude Leibovici
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