$$ \lim_{x\to \infty} \frac{x^{2}+bx+c}{x-n}=\infty $$
recalling the trinomial of the form:
$$ x^{2}+bx+c=(x+n)(x+m) $$
for some $n$ and $m$ such that
$$ m+n=c $$ $$ m\cdot n=b $$
$$ \lim_{x\to \infty} \frac{(x+n)(x+m)}{x-n} $$
And I don't know how to simplify
$$ \lim_{x\to \infty} \frac{x+n}{x-n}\cdot (x+m) $$
Well, you don't need these constants; in fact, you have$$\lim_{x\to +\infty} \frac{x^{2}+bx+c}{x-n}=\lim_{x\to +\infty}x\cdot\frac{1+\frac{b}{x}+\frac{c}{x^2}}{1-\frac{n}{x}}$$
because you can multiply by $\frac{1}{x^2}$ both the numerator and denominator and obtain this form. So, you know that
$$\lim_{x\to +\infty}x=+\infty$$ and
$$\lim_{x\to +\infty}\frac{1+\frac{b}{x}+\frac{c}{x^2}}{1-\frac{n}{x}}=1$$
So thanks to how limits work, you are done. Clearly the same result holds for $-\infty$.
In any polynomial, the leading term dominates the others, and in products or quotients, you can ignore the other terms. Hence,
$$\lim_{x\to\infty}\dfrac{P(x)}{Q(x)}=\lim_{x\to\infty}\dfrac{p_mx^m}{q_nx^n}=\lim_{x\to\infty}\dfrac{p_m}{q_n}x^{m-n}.$$
When $m<n$, the limit is $0$; when $m=n$, $\dfrac{p_m}{q_n}$, and when $m>n$, $\infty$ with the sign of $\dfrac{p_m}{q_n}$.
