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Theorem 1 - Let $f : X \rightarrow \mathbb{R}$ be a function such that $X \subset \mathbb{R}$. If $f$ approaches $l$ near $a$, and $f$ approaches m near a, then $l = m$.

This is from Spivak Calculus, 4e.

I have seen similar question relating to the proof here, where it was asked why we choose $\delta = min(\delta_1, \delta_2)$.

The question remains, why do we choose $|l-m| \over 2$ as our $\epsilon$? I understand that we need to choose an $\epsilon$ to show the falsehood of the statement if we assume $l$ does not equal $m$.

The online version of the textbook is here. The proof can be found by inputting pages 112-113 in the search bar.

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Because the aim is to get $$|l-m|\leq\tfrac12\,|l-m|,$$ thus showing that $l-m=0$. The $\tfrac12$ is irrelevant, any other $t\in(0,1)$ would do.

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  • $\begingroup$ Is this because we are showing a contradiction and $|l - m|$ cannot be less than $1 \over 2$ of itself? $\endgroup$ – NinetyNines Dec 15 '19 at 23:16
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    $\begingroup$ There's no need to think about contradictions. The number $|l-m|$ satisfies the inequality. The only number satisfying that inequality is zero. $\endgroup$ – Martin Argerami Dec 15 '19 at 23:17
  • $\begingroup$ The dissociation was that I wasn't thinking of $|l-m|$ as a number. Thanks for the clarity!! $\endgroup$ – NinetyNines Dec 15 '19 at 23:18
  • $\begingroup$ Actually, I believe the contradiction is that we get $|\ell - m|<2\dfrac{|\ell-m|}2$. $\endgroup$ – Ted Shifrin Dec 16 '19 at 1:24
  • $\begingroup$ @Ted: yes, that depends on how you set up the proof, I guess. If you start with $\varepsilon=|l-m|/2$, then I think one has to go for the contradiction. If instead you take something like $|l-m|/3$, then you get an inequality like $|l-m|\leq 2|l-m|/3$ which allows you to deduce that $|l-m|=0$ directly without appealing to a contradiction argument. $\endgroup$ – Martin Argerami Dec 16 '19 at 2:44
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Suppose both limits exist and that they are different. Thus, given $\epsilon > 0$, there are $\delta_{1} > 0$ and $\delta_{2} > 0$ such that \begin{align*} \begin{cases} 0 < |x - a| < \delta_{1}\\\\ 0 < |x - a| < \delta_{2} \end{cases} \Longrightarrow \begin{cases} |f(x) - m| < \epsilon\\\\ |f(x) - l| < \epsilon \end{cases} \end{align*}

Hence, if we take $\delta = \min\{\delta_{1},\delta_{2}\}$ and $\displaystyle \epsilon = \frac{|l - m|}{2}$, we get the following contradiction: \begin{align*} 0 < |x - a| < \delta \Longrightarrow |l - m| \leq |f(x) - l| + |f(x) - m| < 2\epsilon = |l - m| \end{align*}

which concludes the proof. Hope this helps.

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  • $\begingroup$ Why is it necessary to show that $|l-m| < |l-m|$? Is it because this is a contradiction that $|l-m|$ cannot be strictly less than itself? $\endgroup$ – NinetyNines Dec 15 '19 at 23:17
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    $\begingroup$ That's the point. We have shown that $0 < 0$, which is a contradiction. $\endgroup$ – user1337 Dec 15 '19 at 23:18
  • $\begingroup$ This makes so much sense now!! $\endgroup$ – NinetyNines Dec 15 '19 at 23:19
  • $\begingroup$ Glad to help :) $\endgroup$ – user1337 Dec 16 '19 at 1:56

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