For \(f(x)=\sqrt x\), with limit \(2\) at \(x=4\) and \(\varepsilon=0.1\), what is the largest \(\delta\) with \(0<|x-4|<\delta\Rightarrow|f(x)-2|<0.1\)?
\(|\sqrt x-2|<0.1\) is equivalent to \(3.61<x<4.41\). The distance to \(4\) is \(0.39\) on the left and \(0.41\) on the right; the valid \(\delta\) is the smaller, \(0.39\).
The limit of \(f\) as \(x\) approaches \(a\) is \(L\) if \(f(x)\) gets as close to \(L\) as we like when \(x\) is taken sufficiently close to \(a\), regardless of what happens at \(x=a\).