Complete the \(\delta\)-\(\varepsilon\) definition of a limit: Given a function \(f: \mathbb{R} \to \mathbb{R}\) and \(a, L \in \mathbb{R}\), \(\lim_{x \to a} f(x) = L\) means …
Using the definition, prove that \(\lim_{x \to 1} \frac{1}{1 + x^2} = \frac{1}{2}\).
Solution
The statement \[ \lim_{x\to a}f(x)=L \] means that for every \(\varepsilon>0\) there exists \(\delta>0\) such that \[ 0<|x-a|<\delta\implies |f(x)-L|<\varepsilon. \]
For \(f(x)=1/(1+x^2)\) at \(a=1\), \[ \left|\frac1{1+x^2}-\frac12\right| =\frac{|1-x^2|}{2(1+x^2)} =\frac{|x-1||x+1|}{2(1+x^2)}. \] If \(|x-1|<1\), then \(0<x<2\), hence \(|x+1|<3\) and \(1+x^2\ge1\). Therefore \[ \left|\frac1{1+x^2}-\frac12\right|\le \frac32|x-1|. \] Given \(\varepsilon>0\), choose \[ \delta=\min\left\{1,\frac{2\varepsilon}{3}\right\}. \] Then \(0<|x-1|<\delta\) implies the desired error is less than \(\varepsilon\).