证明：
(设x_{1},x_{2}in (0,1),且x_{1}<x_{2})
(|f(x_{1})-f(x_{2})|=|frac{1}{x_{1}}-frac{1}{x_{2}}|quadquadquad(1)\)
(quadquadquadquadquadquad=frac{x_{2}-x_{1}}{x_{1}x_{2}})
(设frac{x_{2}}{x_{1}}=a,因为x_{2}>x_{1},故a>1\)
(可得x_{2}=ax_{1}\)
(取x_{1}=frac{1}{n},(1)式为：frac{(a-1)n}{a}\)
(forall epsilon>0,\)
(forall delta>0,要求|x_{2}-x{1}|<delta\)
(即：(a-1)x_{1}<delta\)
(即:n>frac{a-1}{delta}quadquad(2)\)
(forall epsilon 如果要求(1)式>epsilon\)
(即：frac{(a-1)n}{a}>epsilon\)
(即：n>frac{epsilon a}{a-1}quadquad(3)\)
(因为a只需>1，不妨取a=2\)
(则(2)式变为：n>frac{1}{delta}\)
((3)式变为n>2epsilon\)
(取N=1+max{2epsilon,frac{1}{delta}}即可\)
(即，forall epsilon>0,forall delta>0\)
(取x_{1}=frac{1}{N},x_2{2}=frac{2}{N}\)
(有(1)式=frac{N}{2}>frac{2epsilon}{2}=epsilon\)
(且满足|x_{2}-x_{1}|=frac{1}{N}<frac{1}{frac{1}{delta}}=delta\)
(证毕\)