Prove $\frac{3n^5 + 20n^3 + 7}{2n^5 - 1}$ converges to $\frac{3}{2}$ by definition Let $a_{n+1} = \frac{3n^5 + 20n^3 + 7}{2n^5 - 1}$, $l = \frac{3}{2}$. Prove that $a_n$ converges to $l$.
I used the absolute value of $a_{n} - l$, to obtain $\frac{40n^3 + 17}{4n^5-2}$. I then obtained these inequalities:
$\frac{40n^3 + 17}{4n^5-2} < \frac{40n^3 + 17}{4n^5-2n^5} = \frac{40n^3 + 17}{2n^5} < \frac{40n^3 + 17n^3}{2n^5} = \frac{57n^3}{2n^5} < \frac{57}{2n^2} < \frac{57}{2n}$, which holds for all natural numbers $n\gt1$.
I then did the usual steps letting $N$ be any natural number greater than $\frac{57}{2\epsilon}$. Does this seem correct  ?
Any help ???
 A: Yes you are on the right track. To elaborate: So let $\epsilon$ be any positive number. We show that there is an $n_0$ such that the equation $|a_n - l| \le \epsilon$ for all integers $n \ge n_0$, holds. That suffices to prove convergence of the sequence $\{a_n\}; n \in \mathbb{N} \ $, to $l=\frac{3}{2}$.
Indeed, let $n_0 = \left\lceil \frac{57}{2\epsilon} \right\rceil$. Then on the one hand, As noted in the OP,
$$|a_n- l| < \frac{57}{2n} \quad \forall n \in \mathbb{N}.$$
On the other hand, for any $n \ge n_0$:
$$\frac{57}{2 \epsilon} \le n \implies \frac{2 \epsilon}{57} \ge \frac{1}{n} \implies  \frac{57}{2n} \le \epsilon .$$
Thus in particular, both statements
$$|a_n- l| < \frac{57}{2n} \quad \forall n \in \mathbb{N},$$ and $$\frac{57}{2n} \le \epsilon \quad \forall n \ge n_0 = \left \lceil \frac{57}{2 \epsilon} \right\rceil,$$ hold. Putting these statements together gives
$$|a_n- l| \le \epsilon \quad \forall {\text{ integers }} n \ge n_0=\left\lceil \frac{57}{2\epsilon} \right\rceil.$$ Thus indeed, we've shown that there is an $n_0$ such that the equation $|a_n - l| \le \epsilon$ for all integers $n \ge n_0$ holds, namely $n_0 = \left \lceil \frac{57}{2 \epsilon} \right \rceil$. As noted above, this suffices to establish convergence of the sequence $\{a_n\}; n \in \mathbb{N} \ $,to $l=\frac{3}{2}$.
A: I tried a different way...
Let $\epsilon>0$ and consider the formal definition of limit for sequences. Then, we want to find an $N(\epsilon)$ such that for all $n>N(\epsilon)$,
$$\left|\frac{3n^5+20n^3+7}{2n^5-1}-\frac{3}{2}\right|<\epsilon\tag{*}$$
i.e.,
$$\left|\frac{40n^3+17}{2(2n^5-1)}\right|<\epsilon$$
is satisfied. Assuming, $N(\epsilon)\geq 1$, then $2n^5-1>0$ and then,
$$\frac{40n^3+17}{2(2n^5-1)}<\epsilon\implies \frac{10}{\epsilon n^2}+\frac{17}{4\epsilon n^5}+\frac{1}{2n^5}<1.$$
We may assume that $\frac{10}{\epsilon n^2}<\frac{1}{3}$, $\frac{17}{4\epsilon n^5}<\frac{1}{3}$, $\frac{1}{2n^5}<\frac{1}{3}$ or $n>\sqrt\frac{30}{\epsilon}$, $n>\sqrt[5]\frac{51}{4\epsilon}$, $n>\sqrt[5]\frac{3}{2}$.
Now, let $$N(\epsilon)=\max\left\{1, \sqrt\frac{30}{\epsilon}, \sqrt[5]\frac{51}{4\epsilon}, \sqrt[5]\frac{3}{2} \right\}.$$
Then, $(*)$ is satisfied for all $N>N(\epsilon)$.
A: Little tip: In such cases it is a very good idea make sense of the polynomials in question. Any rational function $p/q$ can be expressed as $r + p'/q$ where the degree of $p'$ is smaller than the degree of $q$. This is obtained by performing a polynomial division of $p/q$, where $r$ is the quotient and $p'$ is the rest. In our case
$$ p/q = 3/2 $$
with a rest of $$20n^3+17/2$$
So we know:
$$ \frac pq = \frac 32 + \frac{20n^3 + 17/2}{2n^5-1} $$
So the only thing to prove is that $\frac{20n^3 + 17/2}{2n^5-1}$ goes to $0$. For this try to prove:
If $p,q$ are polynomials and $\deg p < \deg q$ then $p(x)/q(x) \to 0$ at $x\to\pm\infty$.
This is quite trivial if $q$ is a monomial.
In fact we can prove: If $\deg p > \deg q$ then $p(x)/q(x) \to \pm\infty$ at $x\to\pm\infty$, and if $\deg p = \deg q$ the whole thing converges to the quotient of the leading coefficients (which is your case).
If $q$ is not a monomial (as in our case) the simplest thing to do would be to multiply $p/q$ with $q/l$ where $l$ is the leading term of $q$. It is again trivial to argue that $q/l\to1$ (as $l$ is a monomial) and this by continuity also $p/q = p/q\cdot q/l = p/l$.
If you do not want to use such a thing you can argue that any term of smaller order  will in absolute value become arbitrarily small compared to the leading term. So you can  for sufficiently large $x$ sandwich each polynomial between the leading term multiplied with factors arbitrarily close to $1$. By using this you can easily achieve this result with only the basic definition of the limit.
