Why do we choose $|l - m| \over 2$ as $\epsilon$ in this proof of the Uniqueness of a Limit?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

You have two inequalities here corresponding to the two limits $l$ and $m$. These inequalities can be combined only when they are considered over the same interval(both on the x-axis and the y-axis). And the only way one can do that is by considering an interval of size $|l-m| \over 2$.