[Math] Trouble understanding Proof from chapter 5 of Spivak’s calculus

calculuslimits

I am trying to understand Theorem 1 from Chapter 5 of Spivak's calculus textbook.

Theorem A function cannot approach two limits $l$ and $m$ for a value $a$.

Proof Let's choose $\epsilon$ such that $|f(x) – l| < \epsilon$ and $|f(x) – m| < \epsilon$ then there exists two numbers such that $ 0 < |x-a| < \delta_1$ and $ 0 < |x-a| < \delta_2$ .
Let $\delta :=\min(\delta_1,\delta_2)$

Here is part where I am lost.

Lets pick a particular $\epsilon$ such that $|f(x) – m| < \epsilon$ and $|f(x) – l| < \epsilon$ cannot both be true if $l\neq m$. He chooses, $\epsilon$ to be midpoint between $l$ and $m$:
$$ \epsilon = \frac{|l-m|}{2}.$$

So by the definition of limit if $0 < |x-a| < \delta$, then $|f(x) – l| < \frac{|l-m|}{2}$ and $|f(x) – m| < \frac{|l-m|}{2}$.

\begin{align*}
|l-m| &= |l – f(x) + f(x) -m| \\
&\leq |l- f(x)| + |m- f(x)|\\
&< \frac{|l-m|}{2} + \frac{|l-m|}{2}
\end{align*}
A contradiction.

I am confused by the last part of the proof where he assumes a condition which cannot be true and goes on to prove that it is not true.

Best Answer

Spivak's proof can be rewritten so that it is no longer a proof by contradiction.

Proposition If $f(x)\to \ell$ and $f(x)\to m$ as $x\to a$, then $|\ell-m|<\varepsilon$ for every $\varepsilon>0$. Thus $\ell=m$.

Proof Given $\varepsilon >0$; pick $\delta >0$ such that $|f(x)-\ell|$ and $|f(x)-m|$ are both less then $\varepsilon/2$ for $0<|x-a|<\delta$; exactly as he did, take a minimum of two appropriate deltas. Now we have that $$|\ell-m|\leqslant |\ell-f(x)|+|f(x)-m|<\frac \varepsilon 2+\frac \varepsilon 2=\varepsilon$$

Since $|\ell-m|\geqslant 0$, and $|\ell-m|<\varepsilon$ is true (as we just showed) for any arbitrary $\varepsilon >0$, we see that $\ell=m$. $\blacktriangleleft$

Claim Suppose $x<\varepsilon$ for every $\varepsilon >0$. Then $x\leqslant 0$.

Proof (Contrapositive) Suppose $x>0$. Then $x>\frac x 2>0$, so taking $\varepsilon =x/2$ we see that for some $\varepsilon >0$, $x\geqslant \varepsilon$. $\blacktriangleleft$

Corollary If $x\geqslant 0$, and $x<\varepsilon$ for every $\varepsilon >0$, then $x=0$.

Proof By the claim, $x\leqslant 0$. Since we already have $x\geqslant 0$, $x=0$. $\blacktriangleleft$

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