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Prove that a sequence cannot converge to two different limits

by RoRi

Prove that a sequence cannot converge to two different limits.

Proof. Consider a sequence \{ a_n \} such that

    \[ \lim_{n \to \infty} a_n = L \qquad \text{and} \qquad \lim_{n \to \infty} a_n = L^*. \]

We show that we must have L = L^*. By the definition of the limit we know that for all \varepsilon > 0 there exist a positive integers N_1 and N_2 such that

    \begin{align*} |a_n - L| &< \frac{\varepsilon}{2} & \text{for all } n \geq N_1 \\ |a_n - L^*| &< \frac{\varepsilon}{2} & \text{for all } n \geq N_2. \end{align*}

Let N = \max \{ N_1, N_2 \}, then for all n \geq N and all \varepsilon > 0 we have

    \[ |a_n - L| < \frac{\varepsilon}{2} \qquad \text{and} \qquad |a_n - L^*| < \frac{\varepsilon}{2}. \]

Hence,

    \begin{align*} &&|a_n - L| + |L^* - a_n| &< \varepsilon \\ \implies && |a_n - L + L^* - a_n| &< \varepsilon \\ \implies && |L^* - L| &< \varepsilon \\ \implies && L = L^*. \end{align*}

(The final line follows since if L \neq L^*, then |L^* - L| > 0, so setting \varepsilon = |L - L^*| would contradict that |L^* - L| < \varepsilon for all \varepsilon > 0.) Therefore, a sequence cannot converge to two different limits. \qquad \blacksquare

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