I am trying to prove that the $\displaystyle \lim_{x\rightarrow \infty}f(x)=L$ is unique, if it exists.
Definition: Let $f:S\rightarrow\mathbb R$ be a function, where $\infty$ is a cluster point of $S$. We say that $f(x)$ converges to $L$ as $x$ goes to $\infty$ if there exists $L\in\mathbb R$ such that for every $\varepsilon>0$, there is an $M\in \mathbb R$ such that $$|f(x)-L|<\varepsilon$$ whenever $x\in S$ and $x\geq M$.
Here is what I got so far:
Proof: Let $L_1$ and $L_2$ be two numbers that both satisfy the above definition. Take $\varepsilon>0$ and find $M_1\in \mathbb R$ such that if $x\in S$ and $x\geq M_1$ then $$|f(x)-L_1|<\varepsilon/2.$$ Also find $M_2\in \mathbb R$ such that if $x\in S$ and $x\geq M_2$ then $$|f(x)-L_2|<\varepsilon/2.$$ Now, put $M=\max\{M_1,M_2\}$. Suppose $x\in S$ and $x\geq M$ (such $x$ exists since $\infty$ is a cluster point of $S$). Then $$|L_1-L_2|=|L_1-f(x)+f(x)-L_2|\leq |L_1-f(x)|+ |f(x)-L_2|<\varepsilon/2 + \varepsilon/2 = \varepsilon.$$ As $|L_1-L_2|<\varepsilon$ for arbitrary $\varepsilon>0$. Then $L_1=L_2$. $\blacksquare$
Do you think this argument is correct? Any help will be appreciated!
Thanks in advanced.
