asymptotic tight bounds for quadratic functions
I think I can suggest more easy way. Having $a&gt;0$ in $f(n)=an^2+bn+c$ we can take any $a_1&lt;a$ and then show that $a_1n^2 &lt; an^2+bn+c$ for some $N_1,n&gt;N_1$ , because it is same as $a_1 &lt; a + \frac{bn+c}{n^2}$.
Same we can make for any $a&lt;a_2 \Rightarrow an^2+bn+c &lt; a_2n^2$ for some $N_2,n&gt;N_2$ because it is same as $a + \frac{bn+c}{n^2}&lt; a_2$.
At last we will have $a_1n^2 &lt;an^2+bn+c &lt; a_2n^2$.