$$\def \C {\mathbb{C}}\def \R {\mathbb{R}}\def \N {\mathbb{N}}\def \Z {\mathbb{Z}}\def \Q {\mathbb{Q}}\def\defn#1{{\bf{#1}}}$$

Note that in the definition of convergence that the definite article is used for the limit, i.e., we have ‘*the* limit’ rather than ‘*a* limit’. A priori there is no reason to expect that a limit like that in the definition is unique. Well, it is and we prove that fact now.

First we need a lemma which should be intuitively obvious: if the difference between two numbers is less than all positive real numbers, then that difference is in fact zero and the numbers are equal.

**Lemma**

Suppose that $x$ and $y$ are real numbers such that $|x-y|\lt \varepsilon $ for all $\varepsilon >0$. Then $x=y$.

**Proof**

Suppose $x\neq y$. Then $|x-y|=\widetilde{\varepsilon}>0$ for some $\widetilde{\varepsilon}\in \R$. But then $|x-y|\lt \varepsilon $ is not true for $\varepsilon =\widetilde{\varepsilon}$ and this is a contradiction.

▢

Now for the observation that limits are unique.

**Theorem (Limits are unique)**

Suppose that $a_n \to a $ and $a_n \to a’ $. Then $a=a’$.

**Proof**

Let $\varepsilon >0$ be an arbitrary real number. Then there exists integers $N$ and $N’$ such that

\[

n>N \Longrightarrow |a_n-a|\lt \frac{\varepsilon }{2}

\]

and

\[

n>N’ \Longrightarrow |a_n-a’|\lt \frac{\varepsilon }{2} .

\]

(If the division by two is confusing, then see the remark following the proof.)

Hence, if $n>\max \{ N, N’ \} $, then

\begin{eqnarray*}

|a-a’|

&=& |a – a_n+ a_n -a’ | \\

&\leq & | a -a_n | + |a_n -a’| , {\text{ by the triangle inequality}}, \\

&=& |a_n -a| + |a_n -a’ | \\

&\lt & \frac{\varepsilon }{2} + \frac{\varepsilon }{2} \\

&=& \varepsilon .

\end{eqnarray*}

Since $\varepsilon $ was arbitrary, by the preceding lemma we deduce that $a=a’$.

▢

**Remark**

The use of $\varepsilon /2$ in the proof is confusing for many students. The objection is that the definition of limit says that given $\varepsilon >0$, there exists an $N$ such that $n>N\Longrightarrow |a_n-a|\lt \varepsilon $. Why can we divide the $\varepsilon $ by $2$?

Well, imagine that you have a machine that takes $\varepsilon$’s and produces $N$’s. Now suppose that someone gives you an $\varepsilon $. You can put $\varepsilon /2$ into the machine (instead of $\varepsilon$) and out will come an $N$ with $n>N\Longrightarrow |a_n-a|\lt \varepsilon /2$. In other words we have the situation that given $\varepsilon >0$, there exists $N$ such that $n>N\Longrightarrow |a_n-a|\lt \varepsilon /2$.

In fact, for a convergent sequence, given any $\varepsilon $ we can find $N$ such that $n>N\Longrightarrow |a_n-a|\lt \varepsilon /3$, or less than $\varepsilon /4$, or $7\varepsilon $, or $\sqrt{\varepsilon}$, or $\varepsilon ^{18}/25000$ or whatever.