For the definitions of sequences and what it means to for a sequence to converge, see this, for subsequences see this.
For the “show the limit” template, see this.
Formal Proof
Since \(\lim a_n = L\), then by definition this means that there exists some \(N_1 \in \mathbb{N}\) such that for all \(n \geq N_1\), we must have
Similarly there exists some \(N_2 \in \mathbb{N}\) such that for all \(n \geq N_2\), we must have
Let \(N = \max(N_1,N_2)\). Then for all \(n \geq N\), we have
Let \(m = n + 1\). Then, whenever \(m \geq N + 1\),
Observe now that
We can now apply the triangle inequality along with the observation that for any \(\epsilon > 0\), since \(a_m \rightarrow L\) and \(a_m \rightarrow M\), there exists an \(N\) such that for \(m \geq N\)
Since \(\epsilon\) was arbitrary, this implies that \(\big\lvert L - M \big\rvert = 0\), and hence \(L = M\) as we wanted to show. \(\blacksquare\)
References