<s>[INST] Find the limit given that $f(1)=1$, $f(x+y)=f(x)+f(y)+2xy$ and $f\left(\frac{1}{x}\right)=\frac{f(x)}{x^4}$ Given the function $f: \mathbb{R}\to \mathbb{R}$ satisfies:
\begin{cases}
f(1)=1 \\
f(x+y)=f(x)+f(y)+2xy,\quad&\forall x,y \in \mathbb{R}\\
f\left(\frac{1}{x}\right)=\frac{f(x)}{x^4},&\forall x\neq0
\tag{1}\label{eqn1}
\end{cases}
Then find the limit of:
\begin{align}
L=\lim_{x \to 0} \frac{\frac{1}{e^{2f(x)}}-\sqrt[3]{1+f(x)}}{\ln\bigl(1+f(x)\bigr)}
\end{align}
My attempts
I can easily guess that $f(x)=x^2$, which totally satisfies \eqref{eqn1}. What I did was that I set $g(x)=f(x)-x^2$, then I got:
\begin{cases}
g(x)+g(y)=g(x+y)\\
g\left(\frac{1}{x}\right)x^4=g(x)
\tag{2}\label{eqn2}
\end{cases}
From \eqref{eqn1}, substitute $x=y=0$ we have $f(0)=0$
Now $g(1)=g(0)=0$, then I substitute $y=1-x$ in \eqref{eqn2} and got that
\begin{align}
g\left(\frac{1}{x}\right)x^4+g\left(\frac{1}{1-x}\right)(1-x)^4=0
\end{align}
Because the values of either $\frac1x$ or $\frac1{1-x}$ must lie between $0$ and $1$, I suspect that we can prove $g \equiv 0$.
Even if I have $f$ then, I'm still clueless about how to calculate the limit.
Any idea or solution or suggestion on the tools?
Any help is appreciated!
 [/INST] The easiest way to go is Taylor expansion.

Then the limit becomes
$$L=\lim_{x\to 0} \frac{e^{-2x^2}-\sqrt[3]{1+x²}}{\ln(1+x^2) } $$
$$=\lim_{x\to 0} \frac{(1-2x^2+2x^4-\cdots +\cdots )-(1+\frac{x^2}{3}-\frac{x^4}{9}+\cdots -\cdots)}{x^2-\frac{x^4}{2}+\frac{x^6}{3}+\cdots} $$
Simplifying this gives
$$L=\lim_{x\to 0} \frac{-\frac{7x^2}{3}+\frac{19x^4}{9}-\cdots +\cdots }{x^2-\frac{x^4}{2}+\frac{x^6}{3}+\cdots} $$
Dividing both numerator and denominator by $x^2$ gives the result
$$L=-\frac{7}{3}$$
</s>