Proofs! Let f(x) = 2x^2 -1 Prove directly from the definition that f(x) is continuous everywhere. Then f(x) is continuous at a iff for every ε0, ∃ δ0 such that |x−a||f(x)−f(a)|

Question

Proofs! Let f(x) = 2x^2 -1 Prove directly from the definition that f(x) is continuous everywhere. Then f(x) is continuous at a iff for every ε>0, ∃ δ>0 such that |x−a|<δ --->|f(x)−f(a)|<ε |f(x) - f(a)| < € |2x^2 -1 - 2a^2 -1| = |2x^2 -2a^2| = 2|x^2-a^2| < € okay so |x^2-a^2| < €/2 I'm pretty sure I'm missing something and need |x^2-a^2| simplified more, I don't know what I can do though.

zarkon · Answer

difference of squares is factorable

anonymous · Answer

I'm not sure how having |(x-a)(x+a)| would help? I could take out the (x+a)|x-a| but is δ allowed to have x and a terms in it? so |x-a|< €/2(x + a) ?

zarkon · Answer

you need to put another restriction on $\delta$.  Something like $\delta<1$

that would put a bound on |x+a|

then your final delta will be the minimum of 1 and some function of epsilon

zarkon · Answer

http://www.math.binghamton.edu/grads/kjones/courses/m221fall07/docs/quadratics.pdf