Using Weierstrass' definition of continuity it is easy to see that [tex]f(x) = x^2[/tex] or [tex]-x^2[/tex] is continuous at [tex]x=0[/tex].
http://en.wikipedia.org/wiki/Continuous ... _functionsUnder this definition all we are required to do to show continuity at [tex]x=0[/tex] is find a function [tex]\delta(\epsilon)>0[/tex] (for [tex]\epsilon>0[/tex]) such that [tex]|x|<\delta(\epsilon)[/tex] implies [tex]x^2<\epsilon[/tex] (we know that [tex]|f(x)-f(0)|=x^2[/tex]). One such function is [tex]\delta(\epsilon) = \sqrt(\epsilon)[/tex] (it is easy to check that this works).
Hope this helped,
R. Baber.