Uniqueness of Limits

Theorem

A function cannot approach two different limits near a.

In other words, if f approaches I near a, and f approaches m near a, then I = m.

Proof

Since $f$ approaches $l$ near $a$ , we know that for any $\epsilon > 0$ there is some number $\delta_1 > 0$ such that, for all $x$

if \quad 0 < |x - a| < \delta_1,\quad then \quad |f(x)-l| < \epsilon.

We also know, since $f$ approaches $m$ near $a$ , that there is some $\delta_2 > 0$ such that, for all $x$ ,

if \quad 0 < |x - a| < \delta_2,\quad then \quad |f(x)-m| < \epsilon.

We have had to use two numbers, $\delta_1$ and $\delta_2$ , since there is no guarantee that the $\delta$ which works in one definition will work in the other.

But, in fact, it is now easy to conclude that for any $\epsilon > 0$ there is some $\delta > 0$ such that, for all $x$ ,

\begin{align} if \quad 0 < |x — a| < \delta, \\ then \quad |f(x) — l| < \epsilon \\ and \quad |f(x) — m| < \epsilon \end{align}

we simply choose

\delta = min(\delta_1, \delta_2)

To complete the proof we just have to pick a particular $\epsilon > 0$ for which the two conditions

|f(x) - l| < \epsilon \quad and \quad |f(x) - m| < \epsilon

cannot both hold, if $l \neq m$ .

The proper choice is suggested by Figure 16. theorem-uniqueness-of_limits-figure-16

If $l \neq m$ , so that $|l-m|>0$ , we can choose $|l-m|/2$ as our $\epsilon$ . It follows that there is a $\delta > 0$ such that, for all $x$ ,

if \quad 0 < |x-a| < \delta,

then \quad |f(x) - l| < \frac{|l-m|}{2}

and \quad |f(x) - m| < \frac{|l-m|}{2}

This implies that for $0 < |x — a | < \delta$ we have

\begin{aligned} |l-m| &= |l - f(x) + f(x) - m| \quad\quad(1)\\ &\leq |l - f(x)| + |f(x) - m| \quad\quad(2)\\ &< \frac{|l-m|}{2} + \frac{|l-m|}{2} \quad\quad(3)\\ &= |l-m| \quad\quad(4) \end{aligned}

a contradiction.

Reference Links:

Step (1) -> (2): Theorem: The Triangle Inequality
Step (2) -> (3): $|a-b| = |b-a|$ because： $|a-b| = |-(b-a)| = |b-a|$ , 因为任何数与其相反数的绝对值都相等

Reference: Calculus Micheal Spivak. 5. Limits

Uniqueness of Limits

Theorem

Proof

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