elementary deduction on limit of sequence Let $(a_n)$ be a convergent sequence and $M$ a real number such that $a_n ≤ M$ for each $n$. Using the previous question, or otherwise, prove that $\lim_{n\to \infty}a_n≤M$.
I tried the "version" where $a_n > M$ and was able to arrived at a solution but this one seems like a tough nut to crack!
 A: For every $\epsilon>0$ there is $N$ such that $|\lim_na_n-a_n|<\epsilon$.
Then $\lim_na_n-a_n<\epsilon$, for $n>N$.
We deduce that $\lim_na_n<a_n+\epsilon\leq M+\epsilon$.
Since we have obtained that for all $\epsilon>0$, $\lim_na_n<M+\epsilon$, therefore $\lim_na_n\leq M$.
A: Let $L$ be the limit of $(a_n)$, and $\varepsilon > 0$. Then, for all $n$ greater than a certain $n_0$, we have: $$\begin{align}-\varepsilon < L-a_n <\varepsilon \\ \Rightarrow L<a_n+\varepsilon\end{align}$$
Since for any $\varepsilon$ there exists an $a_n$ so that the above inequality is satisfied, and each of these $a_n$ is bounded by $M$, we obtain that, for all $\varepsilon > 0$ $$L<M+\varepsilon$$
This means that $L\leq M$ (to prove this, assume $L>M$ and arrive at a contradiction).
Edit: To adress the comment, the first line in the ones you ask about is the definition of $L$ being the limit. Really, the double inequality is just shorthand for the simultaneous inequalities: $$-\varepsilon < L-a_n \\ \text{and} \\L-a_n < \varepsilon$$
Then we just take the second inequality, and add $a_n$ on both sides to get the second line.
