Proving that a convergent sequence has a unique limit

Solution 1:

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.

Solution 2:

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!

Solution 3:

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 $$|x-y|=0~~~\Rightarrow~~~ x = y$$