Consider \[ f(x)=\frac{x}{1-x^2},\qquad x\in(0,1). \]
- Using only the \(\varepsilon\)–\(\delta\) definition, prove that \(f\) is continuous on \((0,1)\).
- Is \(f\) uniformly continuous on \((0,1)\)? Justify your answer.
Solution
1. Prove continuity at an arbitrary \(x_0\in(0,1)\) directly from the definition.
Fix \(x_0\in(0,1)\) and \(\varepsilon>0\). For \(x\in(0,1)\), \[ \begin{aligned} |f(x)-f(x_0)| &=\left|\frac{x}{1-x^2}-\frac{x_0}{1-x_0^2}\right|\\ &=\frac{|x-x_0|\,|1+x x_0|}{(1-x^2)(1-x_0^2)}. \end{aligned} \] Choose \[ \delta :=\min\left\{\frac{1-x_0}{2},\frac{\varepsilon(1-x_0)^2}{4}\right\}. \] If \(|x-x_0|<\delta\), then \[ x<x_0+\frac{1-x_0}{2}=\frac{1+x_0}{2}, \] so \[ 1-x>\frac{1-x_0}{2}. \] Also \(1+x\ge1\), \(1+x_0\ge1\), and \(|1+x x_0|\le2\). Hence \[ (1-x^2)(1-x_0^2) =(1-x)(1+x)(1-x_0)(1+x_0) \ge \frac{(1-x_0)^2}{2}. \] Therefore \[ |f(x)-f(x_0)| \le \frac{4|x-x_0|}{(1-x_0)^2} <\varepsilon. \] Thus \(f\) is continuous at \(x_0\). Since \(x_0\) was arbitrary, \(f\) is continuous on \((0,1)\).
2. \(f\) is not uniformly continuous on \((0,1)\).
For \(n\ge1\), set \[ x_n=\frac{n}{n+1}, \qquad y_n=\frac{n+1}{n+2}. \] Then \[ |x_n-y_n|=\frac{1}{(n+1)(n+2)}\longrightarrow0. \] On the other hand, \[ f(x_n)=\frac{n(n+1)}{2n+1}, \qquad f(y_n)=\frac{(n+1)(n+2)}{2n+3}, \] so \[ |f(y_n)-f(x_n)| =\frac{2(n+1)^2}{(2n+1)(2n+3)} \longrightarrow\frac12. \] If \(f\) were uniformly continuous, taking \(\varepsilon=1/4\) would force \[ |f(y_n)-f(x_n)|<\frac14 \] for all sufficiently large \(n\), because \(|x_n-y_n|\to0\). This contradicts the displayed limit. Therefore \[ \boxed{f\text{ is not uniformly continuous on }(0,1)}. \]