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

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.

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.

• Is this because we are showing a contradiction and $|l - m|$ cannot be less than $1 \over 2$ of itself? – NinetyNines Dec 15 '19 at 23:16
• There's no need to think about contradictions. The number $|l-m|$ satisfies the inequality. The only number satisfying that inequality is zero. – Martin Argerami Dec 15 '19 at 23:17
• The dissociation was that I wasn't thinking of $|l-m|$ as a number. Thanks for the clarity!! – NinetyNines Dec 15 '19 at 23:18
• Actually, I believe the contradiction is that we get $|\ell - m|<2\dfrac{|\ell-m|}2$. – Ted Shifrin Dec 16 '19 at 1:24
• @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. – Martin Argerami Dec 16 '19 at 2:44

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.

• 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? – NinetyNines Dec 15 '19 at 23:17
• That's the point. We have shown that $0 < 0$, which is a contradiction. – user1337 Dec 15 '19 at 23:18
• This makes so much sense now!! – NinetyNines Dec 15 '19 at 23:19
• Glad to help :) – user1337 Dec 16 '19 at 1:56