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