Find the limit given that $f(1)=1$, $f(x+y)=f(x)+f(y)+2xy$ and $f\left(\frac{1}{x}\right)=\frac{f(x)}{{x}^{4}}$

Salvador Bush
2022-07-12
Answered

Find the limit given that $f(1)=1$, $f(x+y)=f(x)+f(y)+2xy$ and $f\left(\frac{1}{x}\right)=\frac{f(x)}{{x}^{4}}$

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conveneau71

Answered 2022-07-13
Author has **17** answers

Then the limit becomes

$L=\underset{x\to 0}{lim}\frac{{e}^{-2{x}^{2}}-\sqrt[3]{1+x\xb2}}{\mathrm{ln}(1+{x}^{2})}$

$=\underset{x\to 0}{lim}\frac{(1-2{x}^{2}+2{x}^{4}-\cdots +\cdots )-(1+\frac{{x}^{2}}{3}-\frac{{x}^{4}}{9}+\cdots -\cdots )}{{x}^{2}-\frac{{x}^{4}}{2}+\frac{{x}^{6}}{3}+\cdots}$

Simplifying this gives

$L=\underset{x\to 0}{lim}\frac{-\frac{7{x}^{2}}{3}+\frac{19{x}^{4}}{9}-\cdots +\cdots}{{x}^{2}-\frac{{x}^{4}}{2}+\frac{{x}^{6}}{3}+\cdots}$

Dividing both numerator and denominator by ${x}^{2}$ gives the result

$L=-\frac{7}{3}$

$L=\underset{x\to 0}{lim}\frac{{e}^{-2{x}^{2}}-\sqrt[3]{1+x\xb2}}{\mathrm{ln}(1+{x}^{2})}$

$=\underset{x\to 0}{lim}\frac{(1-2{x}^{2}+2{x}^{4}-\cdots +\cdots )-(1+\frac{{x}^{2}}{3}-\frac{{x}^{4}}{9}+\cdots -\cdots )}{{x}^{2}-\frac{{x}^{4}}{2}+\frac{{x}^{6}}{3}+\cdots}$

Simplifying this gives

$L=\underset{x\to 0}{lim}\frac{-\frac{7{x}^{2}}{3}+\frac{19{x}^{4}}{9}-\cdots +\cdots}{{x}^{2}-\frac{{x}^{4}}{2}+\frac{{x}^{6}}{3}+\cdots}$

Dividing both numerator and denominator by ${x}^{2}$ gives the result

$L=-\frac{7}{3}$

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