Prove that if $s_n ≤ b$ for all but finitely many $n$, then $\lim s_n ≤ b$. The question asks me to prove  that if $s_n ≤ b$ for all but finitely many $n$, then $\lim s_n ≤ b$ where $(s_n)$ be a sequence that converges.
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Here is how I did it but im not sure if its entirely correct. I used proof by contradiction.
Suppose $\lim s_n>b$ and $s_n \leq b$ for all but finitely many $n$. Let $S=\lim s_n$. By definition we have for every $n>N$ which implies $|s_n-s|<\epsilon$. Let $\epsilon= b-2s+s_n$.
Now we get $s_n-s<\epsilon=b-2s+s_n$ which simplifies to $s<b=\lim s_n<b$ which is a contradiction. 

Let 
$$S=\{n\in\mathbb N\quad|\quad s_n>b\}$$
This set is finite subset of $\mathbb{N}$ so let $n_0=\max S$ hence we have $s_n\leq b\quad \forall n>n_0$ and we conclude.
