Can a function's continuity be described by a uniform value of delta?

[pivoxa15]

Can the epsilon associated with f(x) be a function of x?

i.e epsilon = delta * (x-2)^2 valid?

 

[learningphysics]

No. It shouldn't be function of x.

 

[pivoxa15]

No. It shouldn't be function of x.

Can it be a function of anything?

 

[arildno]

What are you talking about??
Your question is extremely vague.

 

[Super_Leunam]

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|< \delta)

 

[HallsofIvy]

\epsilon and \delta are constants and, as
Super Leunam said, can't depend on anything. However, in most proofs you are given \epsilon (you can't control its value) and must find a correct \delta for that \epsilon. In that sense, \delta may depend on \epsilon.

 

[Kummer]

Unless by x the poster means a specific point x_0.

So for example if you want to show f(x) is continuous at every single point x_0 and you end up with \delta = x_0\epsilon, then that is okay.

 

[quasar987]

Like Kummer said.

If you want to prove continuity at a point x_0, then your expression for \delta(\epsilon) may depend on x_0 too: \delta(\epsilon, x_0). 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 \delta(\epsilon, x_0, f).

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 \delta(\epsilon, x) can be seen as a function sending x in [a,b] to a such proper delta. And if that \delta(\epsilon, x) 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].