Im asked to prove that the function $$ {a_{n}}={{5n+3}\over{8n+9}} $$
is convergent and find the limit $L$, obviously being $5\over8$
I have to use the definition of a limit
$$
\epsilon>0 \ \ \ \exists \ \ N>0 \ \ \ \ s.t \ \ \ \ \ n\ge N \ \Rightarrow \ |a_{n}-L|<\epsilon
$$
I've tried a bunch of different stuff I knew wasn't going to work. I think I have to get $n$ in terms of $\epsilon$ to show that for every $\epsilon$ there is an $n$ in that range but I'm not quite sure how to go about doing that.
For every $\varepsilon > 0$ we have $$ \bigg| \frac{5n+3}{8n+9} - \frac{5}{8} \bigg| = \frac{21}{64n + 72} < \frac{21}{64n} < \varepsilon $$ if $n \geq \lceil \frac{21}{64\varepsilon} \rceil$.