证明：由于$f''left( x ight)$有界，则存在$M>0$，使得对任意的$x in left( {0, + infty } ight)$，有$left| {f''left( x ight)} ight| le M$
由$fleft( x ight) > 0,f'left( x ight) le 0$及单调有界原理知，极限$lim limits_{x o egin{array}{*{20}{c}}{ + infty }end{array}} fleft( x ight)$存在，不妨设为$a$，则$lim limits_{x o egin{array}{*{20}{c}}{ + infty }end{array}} fleft( x ight) = a$
对任意$h>0$，由$f{Taylor公式}$知，存在$xi in left( {x,x + h} ight)$，使得[fleft( {x + h} ight) = fleft( x ight) + f'left( x ight)h + frac{1}{2}f''left( xi ight){h^2}]
则$f'left( x ight) = frac{{fleft( {x + h} ight) - fleft( x ight)}}{h} - frac{h}{2}f''left( xi ight)$，从而可知[left| {f'left( x ight)} ight| le frac{1}{h}left( {left| {fleft( {x + h} ight) - a} ight| + left| {a - fleft( x ight)} ight|} ight) + frac{h}{2}M]
而对任给$varepsilon > 0$，可取$h=hleft( varepsilon ,M ight)>0$，使得$$frac{h}{2}M < frac{varepsilon }{2}$$
且由$lim limits_{x o egin{array}{*{20}{c}}{ + infty }end{array}} fleft( x ight) = a$知，对任给$varepsilon > 0$，存在$G>0$，使得当$x>G$时，有[frac{1}{h}left( {left| {fleft( {x + h} ight) - a} ight| + left| {a - fleft( x ight)} ight|} ight) < 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 ight)} ight| < frac{varepsilon }{2} + frac{varepsilon }{2} = varepsilon $
从而由极限的定义即证