# Find the limit given that f ( 1 ) = 1 , f ( x + y ) = f ( x )

Find the limit given that $f\left(1\right)=1$, $f\left(x+y\right)=f\left(x\right)+f\left(y\right)+2xy$ and $f\left(\frac{1}{x}\right)=\frac{f\left(x\right)}{{x}^{4}}$
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conveneau71
Then the limit becomes
$L=\underset{x\to 0}{lim}\frac{{e}^{-2{x}^{2}}-\sqrt[3]{1+x²}}{\mathrm{ln}\left(1+{x}^{2}\right)}$
$=\underset{x\to 0}{lim}\frac{\left(1-2{x}^{2}+2{x}^{4}-\cdots +\cdots \right)-\left(1+\frac{{x}^{2}}{3}-\frac{{x}^{4}}{9}+\cdots -\cdots \right)}{{x}^{2}-\frac{{x}^{4}}{2}+\frac{{x}^{6}}{3}+\cdots }$
Simplifying this gives
$L=\underset{x\to 0}{lim}\frac{-\frac{7{x}^{2}}{3}+\frac{19{x}^{4}}{9}-\cdots +\cdots }{{x}^{2}-\frac{{x}^{4}}{2}+\frac{{x}^{6}}{3}+\cdots }$
Dividing both numerator and denominator by ${x}^{2}$ gives the result
$L=-\frac{7}{3}$