若 $\lim\limits_{x\to 0}{f\left( x \right)}=f\left( 0 \right)=1$，$f\left( 2x \right)-f\left( x \right)={{x}^{2}}$，求 $f\left( x \right)$．
【难度】
【出处】
2008年清华大学自主招生暨领军计划试题
【标注】
【答案】
$f\left( x \right)=1+\dfrac{{{x}^{2}}}{3}.$
【解析】
由已知，得\[\begin{split}&f\left( x \right)-f\left( \dfrac{1}{2}x \right)=\dfrac{1}{4}{{x}^{2}}\\&f\left( \dfrac{1}{2}x \right)-f\left( \dfrac{1}{4}x \right)=\dfrac{1}{16}{{x}^{2}}\\&f\left( \dfrac{1}{4}x \right)-f\left( \dfrac{1}{8}x \right)=\dfrac{1}{64}{{x}^{2}}\\&\qquad\qquad\cdots\\& f\left( \dfrac{1}{{{2}^{n-1}}}x \right)-f\left( \dfrac{1}{{{2}^{n}}}x \right)=\dfrac{1}{4^n}x^2\end{split}\]累加，得$$f\left( x \right)-f\left( \dfrac{1}{{{2}^{n}}}x \right)=\dfrac{1}{4}{{x}^{2}}+\dfrac{1}{16}{{x}^{2}}+\dfrac{1}{64}{{x}^{2}}+\cdots +\dfrac{1}{{{4}^{n}}}{{x}^{2}},$$即$$f\left( x \right)=f\left( \frac{1}{{{2}^{n}}}x \right)+\frac{{{x}^{2}}}{3}\left[ 1-{{\left( \frac{1}{4} \right)}^{n}} \right],$$又因为$$\underset{x\to 0}{\mathop{\lim }} f\left( x \right)=f\left( 0 \right)=1,$$所以$$f\left( x \right)=\underset{n\to \infty }{\mathop{\lim }} f\left( \frac{1}{{{2}^{n}}}x \right)+\underset{n\to \infty }{\mathop{\lim }} \frac{{{x}^{2}}}{3}\left[ 1-{{\left( \frac{1}{4} \right)}^{n}} \right]=f\left( 0 \right)+\frac{{{x}^{2}}}{3}=1+\frac{{{x}^{2}}}{3}.$$
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