Let \(f\) be a function defined on an interval \((a,b)\) and let \(c \in (a,b)\). Define what it means to say that \(f\) is continuous at \(c\).
Use this definition to show that \(f(x) = 1/x\) is continuous at \(x = 1\).
Solution
A function \(f:(a,b)\to\mathbb R\) is continuous at \(c\in(a,b)\) if for every \(\varepsilon>0\) there exists \(\delta>0\) such that whenever \(x\in(a,b)\) and \(|x-c|<\delta\), one has \[ |f(x)-f(c)|<\varepsilon. \]
For \(f(x)=1/x\) at \(c=1\), let \(\varepsilon>0\) and choose \[ \delta=\min\left\{\frac12,\frac{\varepsilon}{2}\right\}. \] If \(|x-1|<\delta\), then \(x>1/2\), so \(|x|>1/2\). Hence \[ \left|\frac1x-1\right| =\frac{|x-1|}{|x|} <2|x-1| <2\delta \le\varepsilon. \] Thus \(1/x\) is continuous at \(1\).