Let $$(a_n)_{n=m}^{\infty}$$ be a sequence of reals, and let $$L \ne L'$$ be two distinct real numbers. Then $$(a_n)_{n=m}^{\infty}$$ cannot converge to [...].
If $$L \ne L'$$, then there exists a positive real $$d = dist(L, L') = | L - L'|$$. As $$(a_n)_{n=m}^{\infty}$$ converges to $$L$$, we can choose a real $$0 < \varepsilon < \frac{d}{3}$$ and there exists an integer $$N_1 > m$$ such that $$(a_n)_{n=N_1}^{\infty}$$ is [...] $$L$$. Similarly, for $$L'$$ for the same $$\varepsilon$$, there. exists an integer $$N_2 > m$$ such that $$(a_n)_{n=N_2}^{\infty}$$ is [...] $$L'$$. Thus, there exists an $$M = \max(N_1, N_2)$$ (or just $$M = N_1 + N_2$$, if you don't want to use $$\max$$), such that $$a_k$$ is [...] for all $$k \ge M$$. As we have both $$dist(a_k, L) \le \varepsilon$$ and $$dist(a_k, L') \le \varepsilon$$, this implies, by the [...], that $$dist(L, L') \le 2\varepsilon = \frac{2}{3} dist(L, L')$$, which is a false statement, as the distance is always positive. Thus, it cannot be true that $$(a_n)_{n=m}^{\infty}$$ converges to both $$L$$ and $$L'$$.