Proof of the sequence $(a_n) =\frac{\alpha n^2+ \beta n+\gamma}{an^2+bn+c}$ converges? I'm trying to do a assignment about sequences and got a few problems in this question:
Define $(a_n) =\frac{\alpha n^2+ \beta n+\gamma}{an^2+bn+c}$.
Let $a, b, c, \alpha, \beta, \gamma$ be real constants such that $a, \alpha \neq 0$ and $an^2 + bn + c \neq 0$, $\forall \in \mathbb{N}$. Show that $(a_n)$ converges.
My problem with this question is I think it does converges to $\frac{\alpha}{a}$ and they are both different from zero, by creation. But how can I show that?
Thank you in advance!
 A: HINT
We have that
$$a_n =\frac{\alpha n^2+ \beta n+\gamma}{an^2+bn+c}=\frac{\alpha + \beta \frac1n+\gamma\frac1{n^2}}{a+b\frac 1n+c\frac1{n^2}}$$
which is a standard trick for this kind of limit for ratio of polynomials $a_n=\frac{p(n)}{q(n)}$.
Refer also to the related

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*How to find the limit of a function at infinity
A: Another way to go, using asymptotic analysis: a polynomial is asymptotically equivalent to its leading term, therefore
$$\frac{\alpha n^2+ \beta n+\gamma}{an^2+bn+c}\sim_\infty\frac{\alpha n^2}{an^2}=\frac\alpha a.$$
A: $$\frac{\alpha n^2+ \beta n+\gamma}{an^2+bn+c}-\frac\alpha a=\frac{(a\beta-b\alpha)n+(a\gamma-c\alpha)}{a(an^2+bn+c)}\to0$$
because the degree of the denominator exceeds that of the numerator.

Note that this also works when $\alpha=0$ (but not $a=0$).
A: Let us compute the difference between the generic term of your sequence and its limit, and show that it tends to $0$:
$$r_n=a_n-\frac{\alpha}{a} =\frac{\alpha n^2+ \beta n+\gamma}{an^2+bn+c}-\frac{\alpha}{a}$$
($r_n$ as $n$th residual).
$$=\frac{\alpha n^2+ \beta n+\gamma}{an^2+bn+c}-\frac{\alpha}{a}*\frac{n^2+\tfrac{b}{a}n+\tfrac{c}{a}}{n^2+\tfrac{b}{a}n+\tfrac{c}{a}}$$
$$=\frac{\alpha n^2+ \beta n+\gamma}{an^2+bn+c}-\frac{\alpha(n^2+\tfrac{b}{a}n+\tfrac{c}{a})}{an^2+bn+c}$$
$$=\frac{ (\beta-\tfrac{ \alpha b}{a})n+\gamma-\tfrac{\alpha c}{a}}{an^2+bn+c}\tag{1}$$
(because $a \ne 0$) which indeed converges to $0$ because the degree (at most $1$) of the numerator is less than the degree (2) of the denominator.
Edit: An interest of expression (1) is that you can assert, by studying the sign of the equivalent
$$r_n \approx \underbrace{\frac{ (\beta-\tfrac{ \alpha b}{a})}{a}}_C*\frac{1}{n}$$
that (unless $C=0$), the sequence $(a_n)$ converges asymptotically to its limit $\frac{\alpha}{a}$ while being larger than its limit or smaller than its limit, i.e., having an increasing or decreasing (long term) behavior:
If $C>0$, $(a_n)$ is an asymptotically decreasing sequence; if $C<0$, is an increasing sequence.
