Give an \(\varepsilon\)-\(\delta\) proof that \[ \lim_{x\to2}\frac1{3+x}=\frac15. \]
Solution
Let \(\varepsilon>0\) and choose \(\delta=\min\{1,20\varepsilon\}\). If \(|x-2|<\delta\), then \(1<x<3\), so \(|5(3+x)|>20\). Therefore \[ \left|\frac1{3+x}-\frac15\right| =\left|\frac{2-x}{5(3+x)}\right| <\frac1{20}|x-2| <\varepsilon. \]