91Ó°ÊÓ

A function \(f: \mathbb{R} \rightarrow \mathbb{R}\) is uniformly continuous if, for all \( \epsilon > 0\), there exists a \( \delta > 0\) such that if \(|x - y| < \delta\) for any \(x, y \in \mathbb{R}\), then \( |f(x) - f(y)| < \epsilon\). Our goal is to find such \( \delta \) that fulfills these conditions for our function \(f(x)=1/(1+x^{2})\).

\( |f(x)-f(y)|= \left|\frac{1}{1+x^2}-\frac{1}{1+y^2}\right|=\left|\frac{(1+y^2)-(1+x^2)}{(1+x^2)(1+y^2)}\right|=\left|\frac{y^2-x^2}{(1+x^2)(1+y^2)}\right|= \left|\frac{(y-x)(y+x)}{(1+x^2)(1+y^2)}\right|\leq\left|\frac{2|x||y|}{(1+x^2)(1+y^2)}\right|\). You can use the inequality \(1+x^2 \ge 2|x|\) and \(1+y^2\geq2|y|\) to find that \(\left|\frac{2|x||y|}{(1+x^2)(1+y^2)}\right| \leq \frac{1}{2}|x-y|\) . Thus it follows that if \(|x-y| < \frac{2\epsilon}{1}\), then \(|f(x) - f(y)| < \epsilon \). This proves that the function is uniformly continuous. The function \(f(x)\) is uniformly continuous because for every \(\epsilon>0\), we can find \(\delta=\frac{2\epsilon}{1}\), such that if \(|x-y| < \delta\), then \( |f(x)-f(y)| < \epsilon \) holds true.

So now we have justified and illustrated that the function \( f(x)=1/(1+x^{2})\) meets the definition of uniform continuity as for every \(\epsilon>0\), a \(\delta\) can be found, specifically \(\delta=\frac{2\epsilon}{1}\). Hence, this function is uniformly continuous over \(\mathbb{R}\).