Epsilon-delta definition

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please i am new to math. I donot know exact meanings of epsilon-delta definition. i dont comprehend it. Would Anybody help me. thanks in advance
 

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  • #2
vanhees71
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What particularly are you not understanding? Do you mean the [itex]\epsilon[/itex]-[itex]\delta[/itex] definition of continuity of a function at a point?

In this case it reads:

A function [itex]f:D \rightarrow \mathbb{R}[/itex] is continuous at [itex]x_0 \in D[/itex] iff for any [itex]\epsilon>0[/itex] there exists a [itex]\delta>0[/itex] such that for all [itex]x \in D[/itex] with [itex]|x-x_0|<\delta[/itex]
[tex]|f(x)-f(x_0)|<\epsilon.[/tex]
This just says in a formal way that the graph of the function doesn't jump at [itex](x_0,f(x_0))[/itex].
 
  • #3
pwsnafu
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##0<|x-x_0|<\delta## actually.
 
  • #5
jbunniii
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##0<|x-x_0|<\delta## actually.
There's no reason to exclude the point ##x = x_0##. We trivially have ##|f(x_0) - f(x_0)| < \epsilon## for any ##\epsilon##.
 
  • #6
arildno
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There's no reason to exclude the point ##x = x_0##. We trivially have ##|f(x_0) - f(x_0)| < \epsilon## for any ##\epsilon##.
It sure is a reason.
Otherwise, discontinuous functions would be deprived of limit values at the point of discontinuity. Thus, the limit concept would be conflated with the continuity concept.
Think about it!
:smile:
 
  • #7
jbunniii
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It sure is a reason.
Otherwise, discontinuous functions would be deprived of limit values at the point of discontinuity. Thus, the limit concept would be conflated with the continuity concept.
Think about it!
:smile:
But vanhees71 was giving the definition of continuity at ##x_0##, not the definition of the existence of a limit at ##x_0##. If the function is to be continuous at ##x_0##, then it must be defined at ##x_0## and have the correct value!
 
  • #8
arildno
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But vanhees71 was giving the definition of continuity at ##x_0##, not the definition of the existence of a limit at ##x_0##. If the function is to be continuous at ##x_0##, then it must be defined at ##x_0## and have the correct value!
Hmm..no read it again.
What he posted was the definition in terms of the LIMIT concept in which L=f(x_0).
In particular, he writes d>0
 
  • #9
jbunniii
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Hmm..no read it again.
What he posted was the definition in terms of the LIMIT concept in which L=f(x_0).
In particular, he writes d>0
One more remark and then I'll shut up. :tongue: Here's a discussion from Spivak's Calculus (chapter 6, before Theorem 2) which may clarify:

Spivak said:
If we translate the equation ##\lim_{x \rightarrow a} f(x) = f(a)## according to the definition of limits, we obtain:

For every ##\epsilon > 0## there is a ##\delta > 0## such that, for all ##x##, if ##0 < |x - a| < \delta##, then ##|f(x) - f(a)| < \epsilon##.

But in this case, where the limit is ##f(a)##, the phrase ##0 < |x-a| < \delta## may be changed to the simpler condition ##|x-a| < \delta##, since if ##x = a## it is certainly true that ##|f(x) - f(a)| < \epsilon##.
 
  • #10
arildno
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And I do not see why we should make a special case for the continuity criterion relative to the general criterion for a limit?

Why change a perfectly good criterion for continuity?
 

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