If a number sequence has a limit, then the limit is uniquely determined.

Proof. For an indirect proof, suppose that a sequence $$a_1,\,a_2,\,a_3,\,\ldots$$ has two distinct limits $a$ and $b$ . Thus we must have both $$|a_n\!-\!a| < \frac{|a\!-\!b|}{2} \quad \mbox{for all}\;\; n > \mbox{\,some\;} n_1$$ and $$|a_n\!-\!b| < \frac{|a\!-\!b|}{2} \quad \mbox{for all}\;\; n > \mbox{\,some\;} n_2$$ But when $n$ exceeds the greater of $n_1$ and $n_2$ , we can write $$|a\!-\!b| \;=\; |a\!-\!a_n\!+\!a_n\!-\!b|\;\leqq\; |a\!-\!a_n|+|a_n\!-\!b| \;<\; \frac{|a\!-\!b|}{2}+\frac{|a\!-\!b|}{2} \;=\; |a\!-\!b|.$$ This inequality chain contains an impossibility, whence the antithesis made in the begin is wrong and the assertion is right.