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In definition of limits why can't we have " there exist delta for all epsilon" instead of " for all epsilon there exist delta"

saqib
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  • because it leads to a different notions, and a useless one in fact. – Ittay Weiss May 03 '15 at 22:51
  • aren't those statements equivalent? – SalmonKiller May 03 '15 at 22:52
  • thanks for your response....can you clarify by some example? because the statements "there exist delta for all epsilon" and "for all epsilon there exist delta" seems to be same – saqib May 03 '15 at 22:53
  • Nay. What you propose would imply the function is eventually constant on an interval. – Bernard May 03 '15 at 22:54
  • See 'For any natural number $X$ there exists natural $G$ such that $G>X$.' Quite obviously true, isn't it? See now 'There exists a natural number $G$ such that for any natural $X$, $G>X$.' Does this statement seem equivalent to the previous one...? – CiaPan May 04 '15 at 04:52
  • There is a subtle difference between "there exists Y for all X.." and "there exists Y such that for all X..". The first phrase is ambiguous, and will often mean the same as "for all X there exists Y". Example "there exists a nontrivial solution to every linear homogeneous system with more unknowns than equations". However the second possibility unambiguously means something very different; the "such that" is significant here. Nobody would say "there exists a nontrivial solution such that it solves every linear homogeneous system with more unknowns than equations". Please clarify the question – Marc van Leeuwen May 04 '15 at 09:54

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"For any $\epsilon$, there exists a $\delta$" means that whenever you get an $\epsilon$, it's possible to find some $\delta$ that makes it work.

"There exists a $\delta$ such that for any $\epsilon$" means that there is one, single $\delta$ that works no matter what $\epsilon$ might be.

An analogy might do: Take the statement "For any man, there is a woman that is meant for him". It is not an outrageous statement. Some might contest it, but the idea has been there for centuries, and many people will defend its truth to their dying breath.

The statement "There is a woman, such that for any man, she is meant for him", on the other hand, implies that there is one woman somewhere that is everyone's future wife (poor girl). I think you will have to look hard to find anyone who actually believes this.

Arthur
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But a necessary (though not sufficient) condition for that to happen is that $f$ must be uniformly continuous, and not just continuous; otherwise,$\delta$ is a function of $\epsilon $

Gary.
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Because the idea of the limit is that "we can make $f(x)$ arbitrarily close to $L$, by making $x$ sufficiently close to $a$". In other words, instead of analyzing the domain and seeing what happens in the range, we analyze the range, and try to modify the domain appropriately. So we say "for all $\epsilon$ (Range) there exists a $\delta$ (Domain)" instead of vice versa.

SalmonKiller
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  • thanks for your response ... but that's how it is written in book... what if we analyze another way and try to see that other def is same as the usual one given in books – saqib May 03 '15 at 22:57
  • @saqib That's the way they did it before. But that proved insufficient. Read more about it here: https://en.wikipedia.org/wiki/Limit_of_a_function#History – SalmonKiller May 03 '15 at 23:00
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I'll work a similar concept: Continuity.

If we have in the definition "there exists a delta such that for all epsilon", then a lot of intuitively "continuous" functions (those whose graph can be drawn without raising the pencil) wouldn't be continuous.

For instance, take the identity function $f(x)=x$ defined on $\mathbb R$, then $f$ is continuous at $0$ in the intuitive way but not with the "new" definition since it's not true that there exists a $\delta>0$ such that for every $\varepsilon>0$ (whatever you choose), if $|x|<\delta$ then $|f(x)|=|x|<\varepsilon$ (if such a $\delta$ exists, then take $x=\varepsilon=\delta/2$, we have $|x|<\delta$ but $|f(x)|=|x|\nless \varepsilon$).

Daniel
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  • That's not true. There is an $|x|<\delta$ such that $|f(x)-0|<\epsilon$. In fact, for any $\epsilon$ there exists a $\delta = \epsilon$ that would satisfy that. – SalmonKiller May 03 '15 at 23:01
  • Tell if this statement is true: $\exists \delta_0>0: \forall \varepsilon>0: |x|<\delta_0 \implies |f(x)|<\epsilon$. – Daniel May 03 '15 at 23:03
  • without symbolic logic plz. – SalmonKiller May 03 '15 at 23:03
  • What you wrote is right: For every epsilon there exists a delta, that's the classic (and correct one) definition of continuity. Notice that your $\delta$ depends on $\varepsilon$. What I'm arguing is that there can't exist a FIXED $\delta$ that works for all $\varepsilon$. – Daniel May 03 '15 at 23:05
  • that's the definition of the limit. But you can also reverse it and say that for any $\delta$ there is an $\epsilon$ such that ... provided $L+\delta$ and $L-\delta \in$ Domain $f$. – SalmonKiller May 03 '15 at 23:08
  • @SolidSnake are you trying to say that delta changes when ever we change epsilon as they both are same in this case so delta won't remain fixed ? – saqib May 03 '15 at 23:19
  • @saqib Yes, $\delta$ ${\it may}$ vary as $\varepsilon$ does. In this case, $\delta$ can't stay fixed, I've added an explanation of this fact on my answer. – Daniel May 03 '15 at 23:25
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‘There exists a $\delta>0$ such that for all $\varepsilon>0$, if $\,\lvert x-x_0\rvert <\delta$, then $\,\lvert f(x)-\ell\rvert<\varepsilon$’

implies that $f(x)-\ell=0$, i. e. $f(x)=\ell\,$ as soon as $x\in (x_0-\delta,x_0+\delta)$.

Bernard
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