<s>[INST] Limit without de l'Hôpital: $\lim _{x\to 0\color{red}{\boldsymbol -}}\left(1+x^3\right)^{\frac{1}{\left(x^2+1\right)^4-1}}$ 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.
 [/INST] $$A=\left(1+x^3\right)^{\frac{1}{\left(x^2+1\right)^4-1}}\implies \log(A)=\frac{1}{\left(x^2+1\right)^4-1}\log(1+x^3)$$
$$\log(A)=\frac{x^3-\frac{1}{2}x^6+\frac{1}{3}x^9+O\left(x^{12}\right)}{4 x^2+6 x^4+4 x^6+x^8}$$ Long division
$$\log(A)=\frac{x}{4}-\frac{3 x^3}{8}+O\left(x^{4}\right)$$
$$A=e^{\log(A)}=1+\frac{x}{4}+\frac{x^2}{32}+O\left(x^3\right)$$
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