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Here Salman Khan discusses about the formal definition of limits. He tries to prove that a function $f(x)$ has a limit $L$ if $x$ approaches $c$ by expressing $\delta$ as a function of $\epsilon$. In other words, if $\epsilon$ is given, if we can find the value of $\delta$ so that all the $x$ values that satisfy $|x-c|<\delta$, also satisfy $|f(x)-L|<\epsilon$, the limit exists.

My questions:

  1. Is my description of the formal definition of limits correct?
  2. Can we go the other way around? I mean that if we are given $\delta$, if we can find the value of $\epsilon$ so that all $x$ values that satisfy $|f(x)-L|<\epsilon$, also satisfy $|x-c|<\delta$, can we say that the limit exists?
tryingtobeastoic
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  • @DavidC.Ullrich yep, my bad; I edited it. – tryingtobeastoic Sep 30 '21 at 08:14
  • 2 is wrong. However, if you modify it to this, "if we are given $\delta$, if we can find the value of $\epsilon$ so that all $x$ values in this range $c-\delta<x<c+\delta$ that satisfy $|f(x)-L|<\epsilon$, also satisfy $|x-c|<\delta$, can we say that the limit exists?", then you can see this link – tryingtobeastoic Sep 30 '21 at 13:37
  • A better definition of limit can be written in this way: Let $f(x)$ is a real function that is defined for all values near $a$. We can say the limit exists, that is $$\lim_{x\to a}f(x)=L$$, if for every positive $\epsilon$ a corresponding positive $\delta$ is found such that whenever $0<|x-a|<\delta$, $|f(x)-L|<\epsilon$. – tryingtobeastoic Sep 30 '21 at 13:58

1 Answers1

2
  1. Yes.
  2. No, that will not work. Suppose, say, that $f$ is the sine function, and that $c=0$. Then $\lim_{x\to0}\sin(x)=0$ (by the usual definition). But, given $\delta>0$, there is no $\varepsilon>0$ such that, whenever $|\sin(x)|<\varepsilon$, you have $|x|<\delta$, since $\sin(n\pi)=0$ for every $n\in\Bbb Z$.
José Carlos Santos
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