$\frac{1}{x}在(0,1)上不一致连续的证明$

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