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The theorem is as follows:

$$\exists L\in\mathbb R\ \bigg(\lim_{x \to c}f(x)=L\bigg)\iff\forall\varepsilon>0\ \exists\delta>0\ \forall x_1,x_2\in B_{\delta}(c):|f(x_1)-f(x_2)|<\varepsilon$$

where $B_{\delta}$ is the set $\{x\in \mathscr D_f \ | \ 0<|x-c|<\delta\}$

It is quite easy to prove the implication $LHS\Rightarrow RHS$.

It is the other one I am struggling with( $LHS\Leftarrow RHS$ ).I'd appreciate a sketch of the proof or a hint on how to approach this problem. Thanks.

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    $\begingroup$ Isn't that just the definition of limit of a function? $\endgroup$ – DonAntonio May 17 '14 at 13:05
  • $\begingroup$ @DonAntonio Well, the difference is that it uses just the relationship between two "close" points to justify a limit exists. $\endgroup$ – David May 17 '14 at 13:07
  • $\begingroup$ Oh, I just noticed that now, @David...and the claim is clearly false, of course, as $\;L\;$ isn't even involved in the right hand side of that $\;\iff\;$ implication... $\endgroup$ – DonAntonio May 17 '14 at 13:10
  • $\begingroup$ @DonAntonio What I meant by the LHS is that the limit exists. Maybe I should then rephrase it as $\lim_{x \to c}f(x) \; exists$? $\endgroup$ – David May 17 '14 at 13:13
  • $\begingroup$ @DonAntonio: I guess the right hand side implies limit exists $\endgroup$ – mesel May 17 '14 at 13:13
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Your right hand side is called Cauchy's criterion for (continuous) limits.

A hint for the proof is the following: The right hand side implies that for every sequence $(x_i)$ converging to $c$ (with $x_i \neq c$ for all i), the sequence $f(x_i)$ is Cauchy and hence convergent.

By "mixing" two different sequences $(x_i)$ and $(y_i)$ you can show that the limit is independent of the sequence. Call it L and show that this implies that $f(x)$ converges to $L$ as $x \rightarrow c$.

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  • $\begingroup$ so rather than working with two distinct points in the domain of $f$, i.e. $x_1$ and $x_2$, I should work with sequences? $\endgroup$ – David May 17 '14 at 13:43
  • $\begingroup$ Yes, because for these you already know the Cauchy-Criterion (I hope). $\endgroup$ – PhoemueX May 17 '14 at 13:47
  • $\begingroup$ I do. Thanks for the answer. I think I got the proof. $\endgroup$ – David May 17 '14 at 13:48
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David, some ideas for you:

We assume $\;\lim_{x\to c}f(x)=L\;$ (i.e., the limit exists and equals $\;L\;$), and by definition this means that

$$\forall\,\epsilon>0\;\;\exists\,\delta>0\;\;s.t.\;\;|x-c|<\delta\implies |f(x)-L|<\frac\epsilon2$$

so if $\;x_1\,,\,x_2\;$ are two such values (i.e., $\;|x_i-c|<\delta\,,\,i=1,2\;$) , then we get

$$|f(x_1)-f(x_2)|=|f(x_1)-L+L-f(x_2)|\le|f(x_1)-L|+|f(x_2)-L|<\frac\epsilon2+\frac\epsilon2=\epsilon$$

and we're done.

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  • $\begingroup$ I got this part, it is the other implication I can't prove; thanks though. $\endgroup$ – David May 17 '14 at 13:40
  • $\begingroup$ Well @David, you clearly wrote you can't do the $\;RHS\Longleftarrow LHS\;$ implication: this means you can't do the left-to-right hand implication... $\endgroup$ – DonAntonio May 17 '14 at 13:45
  • $\begingroup$ Oh, yeah, sorry I meant the other way. I must've mixed up the letters. sorry again. I'll correct it right away. $\endgroup$ – David May 17 '14 at 13:47

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