Can the epsilon associated with f(x) be a function of x?
i.e epsilon = delta * (x-2)^2 valid?
No. It shouldn't be function of x.
Can it be a function of anything?
What are you talking about??
Your question is extremely vague.
No! it shouldn't be function of anything because it's an ARBITRARY POSITIVE REAL NUMBER ( as such it must be different from 0). When you do an epsilon -- delta proof you must prove that, given an epsilon , which you don't know anything of, some in/equality holds whenever your variable ( x) satisfies some restriction ( i.e. |x-a|< [tex]\delta[/tex])
[itex]\epsilon[/itex] and [itex]\delta[/itex] are constants and, as
Super Leunam said, can't depend on anything. However, in most proofs you are given [itex]\epsilon[/itex] (you can't control its value) and must find a correct [itex]\delta[/itex] for that [itex]\epsilon[/itex]. In that sense, [itex]\delta[/itex] may depend on [itex]\epsilon[/itex].
Unless by [tex]x[/tex] the poster means a specific point [tex]x_0[/tex].
So for example if you want to show [tex]f(x)[/tex] is continous at every single point [tex]x_0[/tex] and you end up with [tex]\delta = x_0\epsilon[/tex], then that is okay.
Like Kummer said.
If you want to prove continuity at a point x_0, then your expression for [itex]\delta(\epsilon)[/itex] may depend on x_0 too: [itex]\delta(\epsilon, x_0)[/itex]. And sometimes, like in the proof of the chain rule I think, it becomes necessary to specify that a number delta is the delta associated with epsilon, and x_0 and of the function f, and in that end, we shall write [itex]\delta(\epsilon, x_0, f)[/itex].
Moreover, if a function is continuous on [a,b] say, then it means that given epsilon>0, there is a delta for every x in [a,b] and [itex]\delta(\epsilon, x)[/itex] can be seen as a function sending x in [a,b] to a such proper delta. And if that [itex]\delta(\epsilon, x)[/itex] function can be arranged to be constant over [a,b] (i.e. independant of x), this is when we say that f is uniformly continuous on [a,b].
Separate names with a comma.