# The uniqueness of a limit of a function: Spivak's proof

I've been slowly reading Spivak's Calculus, and right now I'm on the chapter about limits (chapter five). I'm having trouble understanding a particular step in Spivak's proof of the uniqueness of a limit of a function.

The theorem states:

A function cannot approach two different limits near $a$. In other words, if $f$ approaches $l$ near $a$, and $f$ approaches $m$ near $a$, then $l=m$.

Here is the part of the proof that I fail to understand:

Assume $l\neq m$. Since $f$ approaches $l$ near $a$, we know that for any $\varepsilon > 0$ there is some some number $\delta_1>0$ such that, for all $x$, if $0<|x-a|< \delta_1$, then $|f(x)-l|<\varepsilon$. We also know, since $f$ approaches $m$ near $a$, there is some $\delta_2>0$ such that, for all $x$, if $0<|x-a|< \delta_2$, then $|f(x)-l|<\varepsilon$. 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 $\varepsilon>0$ there is some $\delta>0$ such that, for all $x$, if $0<|x-a|<\delta$, then $|f(x)-l|<\varepsilon$ and $|f(x)-m|<\varepsilon$. We simply choose $\delta=\min (\delta_1, \delta_2)$.

Why did Spivak conclude that the minimum of the two deltas works as the delta needed to combine the two limit definitions?

• If $0 < |x - a| < \min(\delta_1,\delta_2)$, then necessarily, $0 < |x - a| < \delta_1$ and $0 < |x - a| < \delta_2$, since $\min(\delta_1,\delta_2) \le \delta_1,\delta_2$. – qaphla Aug 7 '14 at 0:41

If $0<|x-a|<\min(\delta _1,\delta _2)$, then $0<|x-a|<\delta _1$ and $0<|x-a|<\delta _2$, so $|f(x)-l|<\epsilon$ and $|f(x)-m|<\epsilon$.