Give an \(\varepsilon\)-\(\delta\) proof that \(\lim_{x \to 2} \frac{1}{x^2 + 1} = \frac{1}{5}\).
Solution
We have \[ \left|\frac1{x^2+1}-\frac15\right| =\frac{|4-x^2|}{5(x^2+1)} =\frac{|x-2||x+2|}{5(x^2+1)}. \] If \(|x-2|<1\), then \(1<x<3\), so \(|x+2|<5\) and \(x^2+1>2\). Hence \[ \left|\frac1{x^2+1}-\frac15\right|<\frac12|x-2|. \] Given \(\varepsilon>0\), choose \[ \delta=\min\{1,2\varepsilon\}. \] Then \(0<|x-2|<\delta\) implies the desired error is \(<\varepsilon\).