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证明：由于$f''\left( x \right)$有界，则存在$M>0$，使得对任意的$x \in \left( {0, + \infty } \right)$，有$\left| {f''\left( x \right)} \right| \le M$

由$f\left( x \right) > 0,f'\left( x \right) \le 0$及单调有界原理知，极限$\lim \limits_{x \to \begin{array}{*{20}{c}}{ + \infty }\end{array}} f\left( x \right)$存在，不妨设为$a$，则$\lim \limits_{x \to \begin{array}{*{20}{c}}{ + \infty }\end{array}} f\left( x \right) = a$

对任意$h>0$，由$\bf{Taylor公式}$知，存在$\xi  \in \left( {x,x + h} \right)$，使得\[f\left( {x + h} \right) = f\left( x \right) + f'\left( x \right)h + \frac{1}{2}f''\left( \xi  \right){h^2}\]

则$f'\left( x \right) = \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} - \frac{h}{2}f''\left( \xi  \right)$，从而可知\[\left| {f'\left( x \right)} \right| \le \frac{1}{h}\left( {\left| {f\left( {x + h} \right) - a} \right| + \left| {a - f\left( x \right)} \right|} \right) + \frac{h}{2}M\]

而对任给$\varepsilon  > 0$，可取$h=h\left( \varepsilon ,M \right)>0$，使得$$\frac{h}{2}M < \frac{\varepsilon }{2}$$

且由$\lim \limits_{x \to \begin{array}{*{20}{c}}{ + \infty }\end{array}} f\left( x \right) = a$知，对任给$\varepsilon  > 0$，存在$G>0$，使得当$x>G$时，有\[\frac{1}{h}\left( {\left| {f\left( {x + h} \right) - a} \right| + \left| {a - f\left( x \right)} \right|} \right) < \frac{\varepsilon }{2}\]

所以对任给$\varepsilon  > 0$，存在$G>0$，使得当$x>G$时，有$\left| {f'\left( x \right)} \right| < \frac{\varepsilon }{2} + \frac{\varepsilon }{2} = \varepsilon $

从而由极限的定义即证