TITLE: Proving that a convergent sequence has a unique limit
QUESTION [13 upvotes]: Is the following method wrong?
Let {$a_n$} be a convergent sequence 
Assume $$ \lim_{n \rightarrow \infty} \{a_n \} = L \text{ and}  \lim_{n \rightarrow \infty} \{a_n \} = M$$
$$L-M = \lim_{n \rightarrow \infty} \{a_n \} - \lim_{n \rightarrow \infty} \{a_n \} = L-L =0$$
$$ \therefore  L=M$$

REPLY [7 votes]: 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 $$|x-y|=0~~~\Rightarrow~~~ x = y$$