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The epsilon–delta limit definition $1$:

A function $f(x)$ from $\mathbf{R}$ to $\mathbf{R}$ has limit $L$ at point $x_0 \in \mathbf{R}$ if: $\bbox[yellow] {\text{for every $\epsilon > 0$ there exists a $\delta > 0$ such that whenever $|x-x_0| < \delta$ then}\ |f(x)-L| < \epsilon}$

Intuitive definition $2$:

A function $f(x)$ from $\mathbf{R}$ to $\mathbf{R}$ has limit $L$ at point $x_0 \in \mathbf{R}$ if: $\bbox[yellow] {\text{I can get $f(x)$ as close as I want to $L$ by choosing $x$ close enough to $x_0$}}$

Can anybody explain in a step-by-step manner how the definition $1$ implies definition $2$?

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    $\begingroup$ The intuitive definition should be: I can get $f(x)$ as close as a want to $L$ as long as I choose $x$ close enough to $x_0$. $\endgroup$ – Tychonoff3000 Jun 13 at 13:06
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    $\begingroup$ The intuitive "definition" is wrong and does not follow from the first definition. $\endgroup$ – StackTD Jun 13 at 13:10
  • $\begingroup$ Ok I have edited $\endgroup$ – Joe Jun 13 at 13:26
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Unfortunately, both definitions are incorrect.

The first definition says, rather, that $f$ is continuous at $x_0$ and $f(x_0)=L.$ Instead, it should read

if for all $\epsilon>0,$ there exists $\delta>0$ such that whenever $0<\left\lvert x-x_0\right\rvert<\delta$ then $\left\lvert f(x)-L\right\rvert<\epsilon.$

Consider, for example, the function $$f(x)=\begin{cases}1 & x=0\\0 & x\ne 0.\end{cases}$$

Readily, $f$ has limit $0$ at $x_0=0,$ but doesn't satisfy the given definition, since for $0<\epsilon<1,$ no appropriate $\delta$ can be found.

Moreover, the second definition is inaccurate, as well, as any constant function $f(x)=L$ demonstrates. There is no way for $f(x)$ to get closer to $L$ as $x$ gets closer to $x_0,$ since $f(x)$ is always equal to $f(x_0)$!

Instead, I tend to think about it this way. Saying that $f$ has limit $L$ at $x_0$ says that we're guaranteed to be able to get $f(x)$ as close to $L$ as we like (that is, within $\epsilon$ for any positive $\epsilon$ we choose), so long as we get $x$ sufficiently close to (within some positive $\delta$ of), but not equal to, $x_0.$

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  • $\begingroup$ I am asking if the definition is satisfied, how would it imply that we can get $f(x)$ as close as we want to $L$ by choosing $x$ close enough to $x_0$? Please explain the inner reasonings step by step. $\endgroup$ – Joe Jun 13 at 13:39
  • $\begingroup$ It doesn't imply that. It simply states it formally. If we want $f(x)$ to be within $\epsilon$ of $L$--that is, if we want $|f(x)-L|<\epsilon$--then we're guaranteed that there is a positive $\delta$ such that for $x$ within $\delta$ of (but not equal to) $x_0$--that is, for $x$ such that $0<|x-x_0|<\delta$--we get what we want. $\endgroup$ – Cameron Buie Jun 13 at 13:44

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