# Why the definition of limit is often written

1. Nov 3, 2011

### mahmoud2011

Why the definition of limit is often written in this form also it can be more easy ?
in Real numbers and Real Analysis by Ethan.D.Bloch : writes the definition :
let I$\subseteq$ℝ be an open interval ,c $\in$I , let f:I-{c} → ℝ be a function and let L$\in$ℝ , L is the limit of f as x goes to c ,

if for any ε>0 , there exists δ>0 such that x $\in$ I-{c} and |x-c| < δ imply |f(x)-L| < ε

some questions concerned here , why he don't write instead of the Bold part this simply

0<|x-c| < δ imply |f(x)-L| < ε

in the first definition does the inequality mean that there is some x satisfy it such that x $\notin$ I ?

Last edited: Nov 3, 2011
2. Nov 3, 2011

### micromass

Staff Emeritus
You can write both things. They are equivalent. I have personally always worked with the second definition (i.e. the one with 0<|x-a|<δ ). But either one is good.

3. Nov 3, 2011

### mahmoud2011

Can I prove that the two definitions are equivalent.

4. Nov 3, 2011

### Stephen Tashi

Supose we are talking about the "space" of all functions on the the interval [0,1]. Do we want $lim_{x \rightarrow 0}f(x)$ to fail to exist just because the set $\{x:0 < |x-0|< d \}$ contains points where $f$ is undefined? If you define the topology of $[0,1]$ so that $[1,\delta)$ is an open set, then Bloch's definition allows the limit to exist.

If you use the usual definition of "open set" (as defined on the whole real number line) then the definitions are equivalent.

As I recall, Johnson and Kiokemeister 4th edition gave a definition of limit that was based on topology. They said something like:

"The limit of f(x) as x approaches a is equal to L" means that for each open interval R containing L there is an open interval D containing a such that f(D-{a}) is a subset of R.

However, as I stated this definition, if looks like a failure of f to be defined at various points in D would not prevent the limit from existing. I don't know whether J&K's exact wording prevented that.

Last edited: Nov 3, 2011
5. Nov 3, 2011

### mahmoud2011

Sorry , But I haven't studied topology yet , I am a self learner , also , I have Munkres' Topology , But I didn't read it , Because I still study some set theory , so please I don't want term Topology .