In class today we were given the sequence $a_n=\frac{5n^2+4}{3n^2+n+1}$ and we had to use the epsilon-N definition to prove that it converges to the limit $\frac{5}{3}$
So far I have done as follows
$|\frac{5n^2+4}{3n^2+n+1}-\frac{5}{3}| <\epsilon$
$|\frac{-5n+7}{3(3n^2+n+1)}| < \epsilon$
$\frac{5n^2+7}{9n^2+3n+3} < \epsilon$
$\frac{5n^2+7}{9n^2+3n+3} < \frac{5n^2+7}{9n^2} < \epsilon$
after this i get stuck because I dont know how to simplify this any further. Any help would be much appreciated.
It's unclear why you go from $-5n+7$ to $5n^2+7$.
For $n>1$, $|-5n+7|=5n-7$, so you have to guarantee $$ \frac{5n-7}{3n^2+n+1}<3\varepsilon $$ Note that $5n-7<5n$ and $3n^2+n+1>3n^2$, so $$ \frac{5n-7}{3n^2+n+1}<\frac{5n}{3n^2}=\frac{5}{3n} $$ which is less than $3\varepsilon$ as soon as $$ n>\frac{5}{9\varepsilon} $$