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.