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.

  • 1
    $\begingroup$ It's a proof by contradiction. You assume that a function can approach two limits (l =/= m) at a. You then show that an absurdity occurs (that |l - m| < |l - m|). That means your assumption is incorrect, so we have to have l = m. $\endgroup$ – Tyler Feb 21 '14 at 2:22

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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The proof from Spivak looks clumsy because of symbolic manipulation. If you try to picture the things graphically the inequalities start to look obvious. Lets look at the following figure $$----l--------p--------m----$$ which represents position of $l, m$ on real line and assume that $l < m$ (if things are other way round we can exchange roles of $l, m$). The number $p$ is lying exactly midway between $l, m$ so that $p = (l + m)/2$. It can be easily found that distance between $p$ and $l$ (or between $p$ and $m$) is $\epsilon = (m - l)/2$.

Now Spivak's argument goes as follows. Since $\lim_{x \to a}f(x) = l$ values of $f(x)$ can be made to lie near to $l$ (near meaning that the distance of these values from $l$ does not exceed $\epsilon$ defined above) when $x$ is very near $a$. This means values of $f(x)$ are so near to $l$ that they lie on the left of point $p$ on the real line (This is obvious on the number line. Since $l$ is to the left of $p$, if we try to get very near to $l$ we will ultimately get to the left of $p$). But the similar argument shows that the values of $f(x)$ can be made to lie very near to $m$ and therefore to the right of $p$. This is not possible simultaneously. Thus we can't have $f(x)$ tending to $l$ as well as to $m$ with $l \neq m$.

In most proofs of calculus involving inequalities it makes sense to look at the inequalities by putting the concerned quantities on the real line and then the argument becomes obvious.

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