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.

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    $\begingroup$ 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 $L-M=0$. $\endgroup$ – Pedro Dec 18 '16 at 5:24
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    $\begingroup$ 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? $\endgroup$ – Wrath CC Jan 9 '17 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 Xn = x and lim Xn = y. Since Xn convergent to x for any € > 0 there is an N1 belong to natural number such that | Xn - x | < €/2 for all n >= N1. Similarly for Xn covergent to y. Take n >= max{N1 ,N2} then | Xn - x | < €/2 and | Xn - y | < €/2 hold. Hence | x - y | <= | x - Xn | + | Xn - y | < €/2 + €/2 = € Since € > 0 was arbitrary We conclude that | x - y | = 0, i.e., x = y


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