Find the limit given that $f(1)=1$, $f(x+y)=f(x)+f(y)+2xy$ and $f\left(\frac{1}{x}\right)=\frac{f(x)}{x^4}$
First to prove $g(x) = 0$: We have $g(0) = 0$, $g(1) = f(1) -1^2 = 0$, so $g(x+1) = g(x)$. Also $g(x) + g(-x) = 0$ so $g(-x) = -g(x)$. We have:
$g(x) = g(x+1) = (x+1)^4 g(\frac{1}{x+1}) = -(x+1)^4 g(-\frac{1}{x+1}) = -(x+1)^4 g(1- \frac{1}{x+1}) = -(x+1)^4 g(\frac{x}{x+1}) = -(x+1)^4 (\frac{x}{x+1})^4 g(\frac{x+1}{x}) = -x^4 g(\frac{x+1}{x}) =-x^4 g(1 + \frac{1}{x}) = -x^4 g( \frac{1}{x}) = -x^4 \frac{1}{x^4} g(x) = -g(x)$
Therefore $g(x) = 0$, $f(x) = x^2$. As for the limit, we can substitute $x^2$ with $y$ and apply L'Hôpital's rule .
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
$$L=-\frac{7}{3}$$
Calculating the given limit is standard, when you prove that $ f ( x ) = x ^ 2 $ for all $ x \in \mathbb R $. Let's focus on the nontrivial part of the problem. Using $ g ( 1 ) = 0 $, $$ g ( x + y ) = g ( x ) + g ( y ) \text , \tag 0 \label 0 $$ and $$ g \left ( \frac 1 x \right ) = \frac { g ( x ) } { x ^ 4 } \text , \tag 1 \label 1 $$ you can see that for any $ x \in \mathbb R \setminus \{ - 1 , 0 \} $, \begin{align*} \frac { g \left ( x ^ 2 \right ) } { x ^ 4 ( x + 1 ) ^ 4 } + \frac { g ( x ) } { x ^ 4 ( x + 1 ) ^ 4 } & = \frac { g \left ( x ^ 2 + x \right ) } { \left ( x ^ 2 + x \right ) ^ 4 } \\ & \stackrel { \eqref {1} } = g \left ( \frac 1 { x ^ 2 + x } \right ) \\ & \stackrel { \eqref {0} } = g \left ( \frac 1 x \right ) - g \left ( \frac 1 { x + 1 } \right ) \\ & \stackrel { \eqref {1} } = \frac { g ( x ) } { x ^ 4 } - \frac { g ( x + 1 ) } { ( x + 1 ) ^ 4 } \\ & \stackrel { \eqref {0} } = \frac { g ( x ) } { x ^ 4 } - \frac { g ( x ) + g ( 1 ) } { ( x + 1 ) ^ 4 } \\ & = \frac { \left ( 4 x ^ 3 + 6 x ^ 2 + 4 x + 1 \right ) g ( x ) } { x ^ 4 ( x + 1 ) ^ 4 } \text , \end{align*} which implies $$ g \left ( x ^ 2 \right ) = 2 x ^ 5 ( x + 1 ) ^ 4 \left ( x ^ 2 + 3 x + 1 \right ) g ( x ) \text . \tag 2 \label 2 $$ Using \eqref{0} and \eqref{2} you get \begin{align*} & & & 2 x ^ 5 ( x + 1 ) ^ 4 \left ( x ^ 2 + 3 x + 1 \right ) g ( x ) + 2 g ( x y ) + 2 y ^ 5 ( y + 1 ) ^ 4 \left ( y ^ 2 + 3 y + 1 \right ) g ( y ) \\ & & \stackrel { \eqref {2} } = & g \left ( x ^ 2 \right ) + 2 g ( x y ) + g \left ( y ^ 2 \right ) \\ & & \stackrel { \eqref {0} } = & g \left ( ( x + y ) ^ 2 \right ) \\ & & \stackrel { \eqref {2} } = & 2 ( x + y ) ^ 5 ( x + y + 1 ) ^ 4 \left ( ( x + y ) ^ 2 + 3 ( x + y ) + 1 \right ) g ( x + y ) \\ & & \stackrel { \eqref {0} } = & 2 ( x + y ) ^ 5 ( x + y + 1 ) ^ 4 \left ( ( x + y ) ^ 2 + 3 ( x + y ) + 1 \right ) \bigl ( g ( x ) + g ( y ) \bigr ) \text , \end{align*} which for $ y = 1 $ shows that $$ \Bigl ( ( x + 1 ) ^ 5 ( x + 2 ) ^ 4 \left ( x ^ 2 + 5 x + 5 \right ) - x ^ 5 ( x + 1 ) ^ 4 \left ( x ^ 2 + 3 x + 1 \right ) - 1 \Bigr ) g ( x ) = 0 \text . \tag 3 \label 3 $$ Therefore, $ g ( x ) = 0 $ for all $ x \in \mathbb R \setminus A $, where $ A $ is a finite set of real numbers consisting of $ - 1 $, $ 0 $, and finitely many roots of the polynomial appearing on the left-hand side of \eqref{3}. For any $ a \in A $, choosing a positive integer $ n $ large enough so that $ a + n > b $ for all $ b \in A $, you can use \eqref{0} to see that $ g ( a ) = 0 $, and hence $ g ( x ) = 0 $ for all $ x \in \mathbb R $, or equivalently $ f ( x ) = x ^ 2 $ for all $ x \in \mathbb R $.