# Epsilon-delta proofs?

1. Aug 12, 2007

### pivoxa15

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

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

2. Aug 12, 2007

### learningphysics

No. It shouldn't be function of x.

3. Aug 12, 2007

### pivoxa15

Can it be a function of anything?

4. Aug 12, 2007

### arildno

5. Aug 12, 2007

### 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|< $$\delta$$)

Last edited: Aug 12, 2007
6. Aug 12, 2007

### 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 $\epsilon$.

7. Aug 12, 2007

### Kummer

Unless by $$x$$ the poster means a specific point $$x_0$$.

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

8. Aug 12, 2007

### 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].

Last edited: Aug 12, 2007