Limit without de l'Hôpital: $\lim _{x\to 0\color{red}{\boldsymbol -}}\left(1+x^3\right)^{\frac{1}{\left(x^2+1\right)^4-1}}$

Question

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.

Answer 1

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

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

Answer 2

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…

Answer 3

$$A=\left(1+x^3\right)^{\frac{1}{\left(x^2+1\right)^4-1}}\implies \log(A)=\frac{1}{\left(x^2+1\right)^4-1}\log(1+x^3)$$ $$\log(A)=\frac{x^3-\frac{1}{2}x^6+\frac{1}{3}x^9+O\left(x^{12}\right)}{4 x^2+6 x^4+4 x^6+x^8}$$ Long division $$\log(A)=\frac{x}{4}-\frac{3 x^3}{8}+O\left(x^{4}\right)$$ $$A=e^{\log(A)}=1+\frac{x}{4}+\frac{x^2}{32}+O\left(x^3\right)$$