# Proving that a convergent sequence has a unique limit

Is the following method wrong?

Let {$$a_n$$} be a convergent sequence

Assume $$\lim_{n \rightarrow \infty} \{a_n \} = L \text{ and} \lim_{n \rightarrow \infty} \{a_n \} = M$$

$$L-M = \lim_{n \rightarrow \infty} \{a_n \} - \lim_{n \rightarrow \infty} \{a_n \} = L-L =0$$

$$\therefore L=M$$

If you have already proved the relevant result about the limit of a sum, or difference, it is OK. But the result about uniqueness of limits that you are trying to prove comes typically quite early, immediately after the definition. So we give a fairly detailed proof.

Suppose to the contrary that $L\ne M$. Let $\epsilon=\frac{|L-M|}{10}$.

There is an $N_1$ such that if $n\gt N_1$ then $|a_n-L|\lt \epsilon$.

There is an $N_2$ such that if $n\gt N_2$ then $|a_n-M|\lt \epsilon$.

Let $N=\max(N_1,N_2)$. If $n\gt N$ then $|a_n-L|\lt \epsilon$ and $|a_n-M|\lt \epsilon$.

But then by the Triangle inequality $|L-M|\le |a_n-L|+|M-a_n|\lt \frac{2}{10}|L-M|$. This is impossible. Hence the assumption $L\ne M$ is false and $L=M$.

Remark: The basic intuition is pretty simple. After a while $a_n$ is very close to $L$. After a while it is very close to $M$. That's not possible. The $\epsilon$ stuff made this geometric intuition arithmetical.

• You said: "If you have already proved the relevant result about the limit of a sum, or difference, it is OK". I disagree. Before proving uniqueness, you can prove only the following version of the relevant result about the limit of a difference: if $\{a_n\}$ has a limit $L$ and $\{b_n\}$ has a limit $M$ then $\{a_n- b_n\}$ has a limit $L-M$. As a consequence, from the assumption, we can conclude that $\{a_n-a_n\}$ has a limit $L-M$. Of course, $0$ is also a limit for $\{a_n-a_n\}=\{0\}$ but if we don't have the uniqueness, we can't conclude $L-M=0$. Commented Dec 18, 2016 at 5:24
• Wouldn't the fact that $\{a_n−a_n\}$ is identically zero force $L−M=0$? Is it not known (prior to uniqueness of limits) that $\lim_{n\rightarrow \infty}0=0$, and the limit is unique in the case of a constant sequence? Commented Jan 9, 2017 at 0:21

Suppose that $L \neq M$. Let $\epsilon = |L - M|/2 > 0$. By hypothesis exists $N_1 \in \mathbb{N}$ such that $$|a_n - L| < \dfrac{|L - M|}{2} \quad \text{if} \quad n \geq N_1$$ By hypothesis, exists $N_2 \in \mathbb{N}$ such that $$|a_n - M| < \dfrac{|L - M|}{2} \quad \text{if} \quad n \geq N_2$$ Let $N = \max\{N_1,N_2\}$. If $n \geq N$, then by the triangle inequality $$|L - M| = |(a_n - L) - (a_n - M)| \le |a_ n - L| + |a_n - M| < 2\cdot \dfrac{|L - M|}{2} = |L - M|$$ This is a contradition!

Suppose $$\lim_{n\to\infty}x_n=x$$ and $$\lim_{n\to\infty}x_n=y$$. Since $$x_n$$ converges to $$x$$, for any $$\epsilon > 0$$ there is an $$N_1\in\mathbb N$$ such that $$|x_n-x|<\epsilon/2~~~~~~\forall n\ge N_1$$ Similarly, for any $$\epsilon > 0$$ there is an $$N_2\in\mathbb N$$ such that $$|x_n-y|<\epsilon/2~~~~~~\forall n\ge N_2$$ Let $$N_0=\max\{N1 ,N2\}$$. Then, $$|x_n-x|<\epsilon/2~~~~and~~~~|x_n-y|<\epsilon/2~~~~~~\forall n\ge N_0$$ Now from triangle inequality, $$|x-y|\le|x-x_n|+|x_n-y|<\epsilon/2+\epsilon/2=\epsilon$$ Since $$\epsilon > 0$$ is arbitrary, we conclude that $$|x-y|=0~~~\Rightarrow~~~ x = y$$