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!

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 $\lim_na_n\leq M$.
