Use the \(\varepsilon\)-\(\delta\) definition to prove that \[ \lim_{x\to2}\left(x+\frac1x\right)=\frac52. \]
Solution
Let \(\varepsilon>0\) and choose \[ \delta=\min\left\{1,\frac{2\varepsilon}{5}\right\}. \] If \(0<|x-2|<\delta\), then \(1<x<3\). Hence \(|x|>1\) and \(|2x-1|<5\). Therefore \[ \left|x+\frac1x-\frac52\right| =\frac{|2x-1|\,|x-2|}{2|x|} <\frac52|x-2| <\frac52\delta \leq\varepsilon. \]