Continuity bounds for the square function ($\delta =\min \{1,\epsilon /(2|a|+1)\}$)

For $f(x)=x^2$ at any real $a$, polynomial factorization givesIf $|x-a|<1$, the triangle inequality yields $|x+a|\le |x-a|+2|a|<2|a|+1$. Therefore the displayed choice of $\delta$ makes $|x-a|<\delta$ imply $|f(x)-f(a)|<\epsilon$. This proves continuity and gives an explicit local error bound, rather than an inference from a drawing.

At $a=1$ and $\epsilon =0.6$, this sufficient choice is $\delta =0.2$. The permitted function inputs satisfy $0.8<x<1.2$; their actual function outputs lie in $(0.64,1.44)$, inside the requested band $(0.4,1.6)$. A sufficient input window need not be the largest possible one.