The limit of a sequence, when it exists must be unique.


For the absolute value function definition and other properties see here.
For the definitions of sequences and what it means to for a sequence to converge, see this.
For the “show the limit” template and an example, see this.

Proof:

Let \((x_n)\) be an arbitrary convergent sequence. Suppose for the sake of contradiction that the limit is not unique so we have \((x_n) \rightarrow a\) and \((x_n) \rightarrow b\) for some numbers \(a\) and \(b\) where \(a \neq b\). Without the loss of generality assume that \(a \geq b\). Now, since \((x_n) \rightarrow a\), then for any \(\epsilon > 0\) there exists some \(N_1 \in \mathbf{N}\) such that

$$ \begin{align*} |x_n - a| < \epsilon. \end{align*} $$

Similarly, for any \(\epsilon > 0\) there exists some \(N_2 \in \mathbf{N}\) such that

$$ \begin{align*} |x_n - b| < \epsilon. \end{align*} $$

The trick is to set \(\epsilon\) to \(\frac{b-a}{2}\). Now, let \(M = \max{\{N_1,N_2\}}\) to guarantee having both inequalities. Therefore now we have:

$$ \begin{align*} |x_n - a| < \frac{b-a}{2} \\ |x_n - b| < \frac{b-a}{2}. \end{align*} $$

Let’s simplify these inequalities. Fo the first one,

$$ \begin{align*} |x_n - a| &< \frac{b-a}{2} \\ - (\frac{b-a}{2}) < x_n - a &< \frac{b-a}{2} \\ - (\frac{b-a}{2}) + a < x_n &< \frac{b-a}{2} + a \\ \frac{a-b}{2} + \frac{2a}{2} < x_n & < \frac{b-a}{2} + \frac{2a}{2} \\ \frac{3a-b}{2} < x_n &< \frac{a+b}{2}. \\ \end{align*} $$

And for the second one

$$ \begin{align*} |x_n - b| &< \frac{b-a}{2} \\ - (\frac{b-a}{2}) < x_n - b &< \frac{b-a}{2} \\ - (\frac{b-a}{2}) + b < x_n &< \frac{b-a}{2} + b \\ \frac{a-b}{2} + \frac{2b}{2} < x_n &< \frac{b-a}{2} + \frac{2b}{2} \\ \frac{a+b}{2} < x_n &< \frac{a-3b}{2}. \\ \end{align*} $$

Notice here that we have both \(x_n < \frac{a+b}{2}\) and \(x_n > \frac{a+b}{2}\) and that’s not possible. Therefore the limit must be unique if it exists. \(\blacksquare\)

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